Complex Analysis Revision Lecture
description
Transcript of Complex Analysis Revision Lecture
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MATH20101 Complex Analysis Revision Lecture
Charles Walkden
Dec 11th 2014
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The exam
I The exam paper contains 8 questions:I 4 on Real Analysis (Section A)I 4 on Complex Analysis (Section B).
I You have to answer 5 questions, with at least 2 from eachsection.
I Answer more than 5 then only your best 5 count (subject toat least 2 from each section).
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The exam
I The exam paper contains 8 questions:I 4 on Real Analysis (Section A)I 4 on Complex Analysis (Section B).
I You have to answer 5 questions, with at least 2 from eachsection.
I Answer more than 5 then only your best 5 count (subject toat least 2 from each section).
-
The exam
I The exam paper contains 8 questions:I 4 on Real Analysis (Section A)I 4 on Complex Analysis (Section B).
I You have to answer 5 questions, with at least 2 from eachsection.
I Answer more than 5 then only your best 5 count (subject toat least 2 from each section).
-
Main topics in the course
I 2: The Cauchy-Riemann TheoremI 3: Power seriesI 4: Contour integration and Cauchys TheoremI 5: Cauchys integral formula and its applicationsI 6: Laurent seriesI 7: Poles, residues, the Cauchy Residue Theorem and its uses.
(6 topics, 4 questions...)
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Main topics in the course
I 2: The Cauchy-Riemann TheoremI 3: Power seriesI 4: Contour integration and Cauchys TheoremI 5: Cauchys integral formula and its applicationsI 6: Laurent seriesI 7: Poles, residues, the Cauchy Residue Theorem and its uses.
(6 topics, 4 questions...)
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2 The Cauchy-Riemann Theorem
Let f : D C.
Theorem (Cauchy Riemann Theorem)
IF f is differentible at z0 THEN the partial derivatives exist andthe Cauchy-Riemann equations hold at z0:
u
x(z0) =
v
y(z0),
u
y(z0) = v
x(z0)
Theorem (Converse to the Cauchy-Riemann Theorem)
IF the partial derivatives exist AND ARE CONTINUOUS at z0and the Cauchy-Riemann equations hold at z0 THEN f isdifferentiable at z0.
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2 The Cauchy-Riemann Theorem
Let f : D C.Theorem (Cauchy Riemann Theorem)
IF f is differentible at z0 THEN the partial derivatives exist andthe Cauchy-Riemann equations hold at z0:
u
x(z0) =
v
y(z0),
u
y(z0) = v
x(z0)
Theorem (Converse to the Cauchy-Riemann Theorem)
IF the partial derivatives exist AND ARE CONTINUOUS at z0and the Cauchy-Riemann equations hold at z0 THEN f isdifferentiable at z0.
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2 The Cauchy-Riemann Theorem
Let f : D C.Theorem (Cauchy Riemann Theorem)
IF f is differentible at z0 THEN the partial derivatives exist andthe Cauchy-Riemann equations hold at z0:
u
x(z0) =
v
y(z0),
u
y(z0) = v
x(z0)
Theorem (Converse to the Cauchy-Riemann Theorem)
IF the partial derivatives exist AND ARE CONTINUOUS at z0and the Cauchy-Riemann equations hold at z0 THEN f isdifferentiable at z0.
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3 Power series
A power series at z0 has the form
n=0
an(z z0)n.
an are the coefficients. Think of this as a function of z .
By changing co-ordinates we look at power series at 0:
n=0
anzn ()
Q: When does (*) converge?A: (*) converges for z such that |z | < R, R= radius ofconvergence.(R = means (*) converges for ALL z C.)
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3 Power series
A power series at z0 has the form
n=0
an(z z0)n.
an are the coefficients. Think of this as a function of z .By changing co-ordinates we look at power series at 0:
n=0
anzn ()
Q: When does (*) converge?A: (*) converges for z such that |z | < R, R= radius ofconvergence.(R = means (*) converges for ALL z C.)
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3 Power series
A power series at z0 has the form
n=0
an(z z0)n.
an are the coefficients. Think of this as a function of z .By changing co-ordinates we look at power series at 0:
n=0
anzn ()
Q: When does (*) converge?
A: (*) converges for z such that |z | < R, R= radius ofconvergence.(R = means (*) converges for ALL z C.)
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3 Power series
A power series at z0 has the form
n=0
an(z z0)n.
an are the coefficients. Think of this as a function of z .By changing co-ordinates we look at power series at 0:
n=0
anzn ()
Q: When does (*) converge?A: (*) converges for z such that |z | < R, R= radius ofconvergence.
(R = means (*) converges for ALL z C.)
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3 Power series
A power series at z0 has the form
n=0
an(z z0)n.
an are the coefficients. Think of this as a function of z .By changing co-ordinates we look at power series at 0:
n=0
anzn ()
Q: When does (*) converge?A: (*) converges for z such that |z | < R, R= radius ofconvergence.(R = means (*) converges for ALL z C.)
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3 Radius of convergenceLet
n=0 anz
n be a power series.
Proposition
SUPPOSE the following limit exists
limn
an+1an = 1R .
THEN the radius of convergence is R. (1/0 =, 1/ = 0.)
Example
Define exp(z) =
n=0zn
n! . Then an = 1/n!.an+1an = n!(n + 1)!
= 1n + 1 0 = 1R as n.So R =. So exp(z) converges for all z C.RemarkIf you get a z in your formula for R then youve gone wrong.
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3 Radius of convergenceLet
n=0 anz
n be a power series.
Proposition
SUPPOSE the following limit exists
limn
an+1an = 1R .
THEN the radius of convergence is R. (1/0 =, 1/ = 0.)
Example
Define exp(z) =
n=0zn
n! . Then an = 1/n!.an+1an = n!(n + 1)!
= 1n + 1 0 = 1R as n.So R =. So exp(z) converges for all z C.RemarkIf you get a z in your formula for R then youve gone wrong.
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3 Radius of convergenceLet
n=0 anz
n be a power series.
Proposition
SUPPOSE the following limit exists
limn
an+1an = 1R .
THEN the radius of convergence is R. (1/0 =, 1/ = 0.)
Example
Define exp(z) =
n=0zn
n! .
Then an = 1/n!.an+1an = n!(n + 1)!
= 1n + 1 0 = 1R as n.So R =. So exp(z) converges for all z C.RemarkIf you get a z in your formula for R then youve gone wrong.
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3 Radius of convergenceLet
n=0 anz
n be a power series.
Proposition
SUPPOSE the following limit exists
limn
an+1an = 1R .
THEN the radius of convergence is R. (1/0 =, 1/ = 0.)
Example
Define exp(z) =
n=0zn
n! . Then an = 1/n!.
an+1an = n!(n + 1)!
= 1n + 1 0 = 1R as n.So R =. So exp(z) converges for all z C.RemarkIf you get a z in your formula for R then youve gone wrong.
-
3 Radius of convergenceLet
n=0 anz
n be a power series.
Proposition
SUPPOSE the following limit exists
limn
an+1an = 1R .
THEN the radius of convergence is R. (1/0 =, 1/ = 0.)
Example
Define exp(z) =
n=0zn
n! . Then an = 1/n!.an+1an = n!(n + 1)!
=
1
n + 1 0 = 1
Ras n.
So R =. So exp(z) converges for all z C.RemarkIf you get a z in your formula for R then youve gone wrong.
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3 Radius of convergenceLet
n=0 anz
n be a power series.
Proposition
SUPPOSE the following limit exists
limn
an+1an = 1R .
THEN the radius of convergence is R. (1/0 =, 1/ = 0.)
Example
Define exp(z) =
n=0zn
n! . Then an = 1/n!.an+1an = n!(n + 1)!
= 1n + 1
0 = 1R
as n.
So R =. So exp(z) converges for all z C.RemarkIf you get a z in your formula for R then youve gone wrong.
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3 Radius of convergenceLet
n=0 anz
n be a power series.
Proposition
SUPPOSE the following limit exists
limn
an+1an = 1R .
THEN the radius of convergence is R. (1/0 =, 1/ = 0.)
Example
Define exp(z) =
n=0zn
n! . Then an = 1/n!.an+1an = n!(n + 1)!
= 1n + 1 0 = 1R as n.
So R =. So exp(z) converges for all z C.RemarkIf you get a z in your formula for R then youve gone wrong.
-
3 Radius of convergenceLet
n=0 anz
n be a power series.
Proposition
SUPPOSE the following limit exists
limn
an+1an = 1R .
THEN the radius of convergence is R. (1/0 =, 1/ = 0.)
Example
Define exp(z) =
n=0zn
n! . Then an = 1/n!.an+1an = n!(n + 1)!
= 1n + 1 0 = 1R as n.So R =. So exp(z) converges for all z C.
RemarkIf you get a z in your formula for R then youve gone wrong.
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3 Radius of convergenceLet
n=0 anz
n be a power series.
Proposition
SUPPOSE the following limit exists
limn
an+1an = 1R .
THEN the radius of convergence is R. (1/0 =, 1/ = 0.)
Example
Define exp(z) =
n=0zn
n! . Then an = 1/n!.an+1an = n!(n + 1)!
= 1n + 1 0 = 1R as n.So R =. So exp(z) converges for all z C.RemarkIf you get a z in your formula for R then youve gone wrong.
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3 Properties of power seriesTheoremLet f (z) =
n=0 anz
n have radius of convergence R. Then f (z) isholomorphic on the disc of convergence {z C | |z | < R} andf (z) =
n=1 nanz
n1.
Example
exp(z) =n=0
zn
n!
= 1 + z +z2
2!+
z3
3!+
z4
4!+ .
exp(z) = 1 + z +z2
2!+
z3
3!+ = exp(z).
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3 Properties of power seriesTheoremLet f (z) =
n=0 anz
n have radius of convergence R. Then f (z) isholomorphic on the disc of convergence {z C | |z | < R} andf (z) =
n=1 nanz
n1.
Example
exp(z) =n=0
zn
n!
= 1 + z +z2
2!+
z3
3!+
z4
4!+ .
exp(z) = 1 + z +z2
2!+
z3
3!+ = exp(z).
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3 Properties of power seriesTheoremLet f (z) =
n=0 anz
n have radius of convergence R. Then f (z) isholomorphic on the disc of convergence {z C | |z | < R} andf (z) =
n=1 nanz
n1.
Example
exp(z) =n=0
zn
n!
= 1 + z +z2
2!+
z3
3!+
z4
4!+ .
exp(z) =
1 + z +z2
2!+
z3
3!+ = exp(z).
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3 Properties of power seriesTheoremLet f (z) =
n=0 anz
n have radius of convergence R. Then f (z) isholomorphic on the disc of convergence {z C | |z | < R} andf (z) =
n=1 nanz
n1.
Example
exp(z) =n=0
zn
n!
= 1 + z +z2
2!+
z3
3!+
z4
4!+ .
exp(z) = 1 + z +z2
2!+
z3
3!+ =
exp(z).
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3 Properties of power seriesTheoremLet f (z) =
n=0 anz
n have radius of convergence R. Then f (z) isholomorphic on the disc of convergence {z C | |z | < R} andf (z) =
n=1 nanz
n1.
Example
exp(z) =n=0
zn
n!
= 1 + z +z2
2!+
z3
3!+
z4
4!+ .
exp(z) = 1 + z +z2
2!+
z3
3!+ = exp(z).
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3 Properties of power series
TheoremLet f (z) =
n=0 anz
n have radius of convergence R. Then f (z) isholomorphic on the disc of convergence {z C | |z | < R} andf (z) =
n=1 nanz
n1.
How did we prove this?
I Step 1: Show that the power series g(z) =
n=1 nanzn1 has
radius of convergence at least R.
I Step 2: Show that f (z) is differentiable and f (z) = g(z).Step 2 is hard (so not examinable).Step 1 is fun!
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3 Properties of power series
TheoremLet f (z) =
n=0 anz
n have radius of convergence R. Then f (z) isholomorphic on the disc of convergence {z C | |z | < R} andf (z) =
n=1 nanz
n1.How did we prove this?
I Step 1: Show that the power series g(z) =
n=1 nanzn1 has
radius of convergence at least R.
I Step 2: Show that f (z) is differentiable and f (z) = g(z).Step 2 is hard (so not examinable).Step 1 is fun!
-
3 Properties of power series
TheoremLet f (z) =
n=0 anz
n have radius of convergence R. Then f (z) isholomorphic on the disc of convergence {z C | |z | < R} andf (z) =
n=1 nanz
n1.How did we prove this?
I Step 1: Show that the power series g(z) =
n=1 nanzn1 has
radius of convergence at least R.
I Step 2: Show that f (z) is differentiable and f (z) = g(z).Step 2 is hard (so not examinable).Step 1 is fun!
-
3 Properties of power series
TheoremLet f (z) =
n=0 anz
n have radius of convergence R. Then f (z) isholomorphic on the disc of convergence {z C | |z | < R} andf (z) =
n=1 nanz
n1.How did we prove this?
I Step 1: Show that the power series g(z) =
n=1 nanzn1 has
radius of convergence at least R.
I Step 2: Show that f (z) is differentiable and f (z) = g(z).
Step 2 is hard (so not examinable).Step 1 is fun!
-
3 Properties of power series
TheoremLet f (z) =
n=0 anz
n have radius of convergence R. Then f (z) isholomorphic on the disc of convergence {z C | |z | < R} andf (z) =
n=1 nanz
n1.How did we prove this?
I Step 1: Show that the power series g(z) =
n=1 nanzn1 has
radius of convergence at least R.
I Step 2: Show that f (z) is differentiable and f (z) = g(z).Step 2 is hard (so not examinable).Step 1 is fun!
-
Want to prove: IF
n=0 anzn has radius of convergence R and
|z | < R THEN n=1 nanzn1 converges absolutely.
Q: Can we use the formula:
n=0 bnzn has radius of convergence
1R = limn
bn+1bn ?A: NO! We dont know this limit exists! (In general, it doesnt.)
Let |z | < R and choose r s.t. |z | < r < R. Then n=0 anrnconverges absolutely.Hence the summands must tend to zero. Hence the summandsmust be bounded: i.e. K > 0 s.t. |anrn| < K n 0.Let q = |z|r (0, 1). Then
|nanzn1| = n|an|z
r
n1 rn1 = n |anrn|r
zr
n1 < Kr
nqn1.
But (fact!)
n=1 nqn1 converges to (1 q)2.
By the comparison test,
n=1 |nanzn1| converges.
-
Want to prove: IF
n=0 anzn has radius of convergence R and
|z | < R THEN n=1 nanzn1 converges absolutely.Q: Can we use the formula:
n=0 bnz
n has radius of convergence1R = limn
bn+1bn ?
A: NO! We dont know this limit exists! (In general, it doesnt.)
Let |z | < R and choose r s.t. |z | < r < R. Then n=0 anrnconverges absolutely.Hence the summands must tend to zero. Hence the summandsmust be bounded: i.e. K > 0 s.t. |anrn| < K n 0.Let q = |z|r (0, 1). Then
|nanzn1| = n|an|z
r
n1 rn1 = n |anrn|r
zr
n1 < Kr
nqn1.
But (fact!)
n=1 nqn1 converges to (1 q)2.
By the comparison test,
n=1 |nanzn1| converges.
-
Want to prove: IF
n=0 anzn has radius of convergence R and
|z | < R THEN n=1 nanzn1 converges absolutely.Q: Can we use the formula:
n=0 bnz
n has radius of convergence1R = limn
bn+1bn ?A: NO! We dont know this limit exists! (In general, it doesnt.)
Let |z | < R and choose r s.t. |z | < r < R. Then n=0 anrnconverges absolutely.Hence the summands must tend to zero. Hence the summandsmust be bounded: i.e. K > 0 s.t. |anrn| < K n 0.Let q = |z|r (0, 1). Then
|nanzn1| = n|an|z
r
n1 rn1 = n |anrn|r
zr
n1 < Kr
nqn1.
But (fact!)
n=1 nqn1 converges to (1 q)2.
By the comparison test,
n=1 |nanzn1| converges.
-
Want to prove: IF
n=0 anzn has radius of convergence R and
|z | < R THEN n=1 nanzn1 converges absolutely.Q: Can we use the formula:
n=0 bnz
n has radius of convergence1R = limn
bn+1bn ?A: NO! We dont know this limit exists! (In general, it doesnt.)
Let |z | < R and choose r s.t. |z | < r < R.
Then
n=0 anrn
converges absolutely.Hence the summands must tend to zero. Hence the summandsmust be bounded: i.e. K > 0 s.t. |anrn| < K n 0.Let q = |z|r (0, 1). Then
|nanzn1| = n|an|z
r
n1 rn1 = n |anrn|r
zr
n1 < Kr
nqn1.
But (fact!)
n=1 nqn1 converges to (1 q)2.
By the comparison test,
n=1 |nanzn1| converges.
-
Want to prove: IF
n=0 anzn has radius of convergence R and
|z | < R THEN n=1 nanzn1 converges absolutely.Q: Can we use the formula:
n=0 bnz
n has radius of convergence1R = limn
bn+1bn ?A: NO! We dont know this limit exists! (In general, it doesnt.)
Let |z | < R and choose r s.t. |z | < r < R. Then n=0 anrnconverges absolutely.
Hence the summands must tend to zero. Hence the summandsmust be bounded: i.e. K > 0 s.t. |anrn| < K n 0.Let q = |z|r (0, 1). Then
|nanzn1| = n|an|z
r
n1 rn1 = n |anrn|r
zr
n1 < Kr
nqn1.
But (fact!)
n=1 nqn1 converges to (1 q)2.
By the comparison test,
n=1 |nanzn1| converges.
-
Want to prove: IF
n=0 anzn has radius of convergence R and
|z | < R THEN n=1 nanzn1 converges absolutely.Q: Can we use the formula:
n=0 bnz
n has radius of convergence1R = limn
bn+1bn ?A: NO! We dont know this limit exists! (In general, it doesnt.)
Let |z | < R and choose r s.t. |z | < r < R. Then n=0 anrnconverges absolutely.Hence the summands must tend to zero. Hence the summandsmust be bounded:
i.e. K > 0 s.t. |anrn| < K n 0.Let q = |z|r (0, 1). Then
|nanzn1| = n|an|z
r
n1 rn1 = n |anrn|r
zr
n1 < Kr
nqn1.
But (fact!)
n=1 nqn1 converges to (1 q)2.
By the comparison test,
n=1 |nanzn1| converges.
-
Want to prove: IF
n=0 anzn has radius of convergence R and
|z | < R THEN n=1 nanzn1 converges absolutely.Q: Can we use the formula:
n=0 bnz
n has radius of convergence1R = limn
bn+1bn ?A: NO! We dont know this limit exists! (In general, it doesnt.)
Let |z | < R and choose r s.t. |z | < r < R. Then n=0 anrnconverges absolutely.Hence the summands must tend to zero. Hence the summandsmust be bounded: i.e. K > 0 s.t. |anrn| < K n 0.
Let q = |z|r (0, 1). Then
|nanzn1| = n|an|z
r
n1 rn1 = n |anrn|r
zr
n1 < Kr
nqn1.
But (fact!)
n=1 nqn1 converges to (1 q)2.
By the comparison test,
n=1 |nanzn1| converges.
-
Want to prove: IF
n=0 anzn has radius of convergence R and
|z | < R THEN n=1 nanzn1 converges absolutely.Q: Can we use the formula:
n=0 bnz
n has radius of convergence1R = limn
bn+1bn ?A: NO! We dont know this limit exists! (In general, it doesnt.)
Let |z | < R and choose r s.t. |z | < r < R. Then n=0 anrnconverges absolutely.Hence the summands must tend to zero. Hence the summandsmust be bounded: i.e. K > 0 s.t. |anrn| < K n 0.Let q = |z|r (0, 1).
Then
|nanzn1| = n|an|z
r
n1 rn1 = n |anrn|r
zr
n1 < Kr
nqn1.
But (fact!)
n=1 nqn1 converges to (1 q)2.
By the comparison test,
n=1 |nanzn1| converges.
-
Want to prove: IF
n=0 anzn has radius of convergence R and
|z | < R THEN n=1 nanzn1 converges absolutely.Q: Can we use the formula:
n=0 bnz
n has radius of convergence1R = limn
bn+1bn ?A: NO! We dont know this limit exists! (In general, it doesnt.)
Let |z | < R and choose r s.t. |z | < r < R. Then n=0 anrnconverges absolutely.Hence the summands must tend to zero. Hence the summandsmust be bounded: i.e. K > 0 s.t. |anrn| < K n 0.Let q = |z|r (0, 1). Then
|nanzn1| = n|an|z
r
n1 rn1 = n |anrn|r
zr
n1 < Kr
nqn1.
But (fact!)
n=1 nqn1 converges to (1 q)2.
By the comparison test,
n=1 |nanzn1| converges.
-
Want to prove: IF
n=0 anzn has radius of convergence R and
|z | < R THEN n=1 nanzn1 converges absolutely.Q: Can we use the formula:
n=0 bnz
n has radius of convergence1R = limn
bn+1bn ?A: NO! We dont know this limit exists! (In general, it doesnt.)
Let |z | < R and choose r s.t. |z | < r < R. Then n=0 anrnconverges absolutely.Hence the summands must tend to zero. Hence the summandsmust be bounded: i.e. K > 0 s.t. |anrn| < K n 0.Let q = |z|r (0, 1). Then
|nanzn1| = n|an|z
r
n1 rn1 = n |anrn|r
zr
n1 < Kr
nqn1.
But (fact!)
n=1 nqn1 converges to (1 q)2.
By the comparison test,
n=1 |nanzn1| converges.
-
Want to prove: IF
n=0 anzn has radius of convergence R and
|z | < R THEN n=1 nanzn1 converges absolutely.Q: Can we use the formula:
n=0 bnz
n has radius of convergence1R = limn
bn+1bn ?A: NO! We dont know this limit exists! (In general, it doesnt.)
Let |z | < R and choose r s.t. |z | < r < R. Then n=0 anrnconverges absolutely.Hence the summands must tend to zero. Hence the summandsmust be bounded: i.e. K > 0 s.t. |anrn| < K n 0.Let q = |z|r (0, 1). Then
|nanzn1| = n|an|z
r
n1 rn1 = n |anrn|r
zr
n1 < Kr
nqn1.
But (fact!)
n=1 nqn1 converges to (1 q)2.
By the comparison test,
n=1 |nanzn1| converges.
-
4 Contour integrationSuppose f : D C, is a path in D.
Let : [a, b] D be a parametrisation of .Definition
f =
ba
f ((t))(t) dt.
Properties of the contour integral:1+2
f =
1
f +
2
f .
f =
f .
(Remember: is the REVERSED PATH. If (t), a t b isa path then (t) is the path (t) = (a + b t), a t b.)
-
4 Contour integrationSuppose f : D C, is a path in D.Let : [a, b] D be a parametrisation of .
Definition
f =
ba
f ((t))(t) dt.
Properties of the contour integral:1+2
f =
1
f +
2
f .
f =
f .
(Remember: is the REVERSED PATH. If (t), a t b isa path then (t) is the path (t) = (a + b t), a t b.)
-
4 Contour integrationSuppose f : D C, is a path in D.Let : [a, b] D be a parametrisation of .Definition
f =
ba
f ((t))(t) dt.
Properties of the contour integral:1+2
f =
1
f +
2
f .
f =
f .
(Remember: is the REVERSED PATH. If (t), a t b isa path then (t) is the path (t) = (a + b t), a t b.)
-
4 Contour integrationSuppose f : D C, is a path in D.Let : [a, b] D be a parametrisation of .Definition
f =
ba
f ((t))(t) dt.
Properties of the contour integral:
1+2
f =
1
f +
2
f .
f =
f .
(Remember: is the REVERSED PATH. If (t), a t b isa path then (t) is the path (t) = (a + b t), a t b.)
-
4 Contour integrationSuppose f : D C, is a path in D.Let : [a, b] D be a parametrisation of .Definition
f =
ba
f ((t))(t) dt.
Properties of the contour integral:1+2
f =
1
f +
2
f .
f =
f .
(Remember: is the REVERSED PATH. If (t), a t b isa path then (t) is the path (t) = (a + b t), a t b.)
-
4 Contour integrationSuppose f : D C, is a path in D.Let : [a, b] D be a parametrisation of .Definition
f =
ba
f ((t))(t) dt.
Properties of the contour integral:1+2
f =
1
f +
2
f .
f =
f .
(Remember: is the REVERSED PATH. If (t), a t b isa path then (t) is the path (t) = (a + b t), a t b.)
-
4 The Fundamental Theorem of Contour Integration
DefinitionSuppose there exists a function F : D C such that F (z) = f (z)for all z D. Call F an anti-derivative of f .Theorem (Fundamental Theorem of Contour Integration)
Suppose f : D C has an anti-derivative F . Suppose is a pathfrom z0 to z1. Then
f = F (z1) F (z0).
Example
f (z) = z2. f has antiderivative F (z) = z3/3.Let be any path from 0 to i .Then
f = F (i) F (0) = i
3
3= i
3.
RemarkMost functions DO NOT have an anti-derivative. Examplef : C \ {0} C, f (z) = 1/z .
-
4 The Fundamental Theorem of Contour IntegrationDefinitionSuppose there exists a function F : D C such that F (z) = f (z)for all z D. Call F an anti-derivative of f .
Theorem (Fundamental Theorem of Contour Integration)
Suppose f : D C has an anti-derivative F . Suppose is a pathfrom z0 to z1. Then
f = F (z1) F (z0).
Example
f (z) = z2. f has antiderivative F (z) = z3/3.Let be any path from 0 to i .Then
f = F (i) F (0) = i
3
3= i
3.
RemarkMost functions DO NOT have an anti-derivative. Examplef : C \ {0} C, f (z) = 1/z .
-
4 The Fundamental Theorem of Contour IntegrationDefinitionSuppose there exists a function F : D C such that F (z) = f (z)for all z D. Call F an anti-derivative of f .Theorem (Fundamental Theorem of Contour Integration)
Suppose f : D C has an anti-derivative F . Suppose is a pathfrom z0 to z1. Then
f = F (z1) F (z0).
Example
f (z) = z2. f has antiderivative F (z) = z3/3.Let be any path from 0 to i .Then
f = F (i) F (0) = i
3
3= i
3.
RemarkMost functions DO NOT have an anti-derivative. Examplef : C \ {0} C, f (z) = 1/z .
-
4 The Fundamental Theorem of Contour IntegrationDefinitionSuppose there exists a function F : D C such that F (z) = f (z)for all z D. Call F an anti-derivative of f .Theorem (Fundamental Theorem of Contour Integration)
Suppose f : D C has an anti-derivative F . Suppose is a pathfrom z0 to z1. Then
f = F (z1) F (z0).
Example
f (z) = z2.
f has antiderivative F (z) = z3/3.Let be any path from 0 to i .Then
f = F (i) F (0) = i
3
3= i
3.
RemarkMost functions DO NOT have an anti-derivative. Examplef : C \ {0} C, f (z) = 1/z .
-
4 The Fundamental Theorem of Contour IntegrationDefinitionSuppose there exists a function F : D C such that F (z) = f (z)for all z D. Call F an anti-derivative of f .Theorem (Fundamental Theorem of Contour Integration)
Suppose f : D C has an anti-derivative F . Suppose is a pathfrom z0 to z1. Then
f = F (z1) F (z0).
Example
f (z) = z2. f has antiderivative F (z) = z3/3.
Let be any path from 0 to i .Then
f = F (i) F (0) = i
3
3= i
3.
RemarkMost functions DO NOT have an anti-derivative. Examplef : C \ {0} C, f (z) = 1/z .
-
4 The Fundamental Theorem of Contour IntegrationDefinitionSuppose there exists a function F : D C such that F (z) = f (z)for all z D. Call F an anti-derivative of f .Theorem (Fundamental Theorem of Contour Integration)
Suppose f : D C has an anti-derivative F . Suppose is a pathfrom z0 to z1. Then
f = F (z1) F (z0).
Example
f (z) = z2. f has antiderivative F (z) = z3/3.Let be any path from 0 to i .
Then
f = F (i) F (0) = i3
3= i
3.
RemarkMost functions DO NOT have an anti-derivative. Examplef : C \ {0} C, f (z) = 1/z .
-
4 The Fundamental Theorem of Contour IntegrationDefinitionSuppose there exists a function F : D C such that F (z) = f (z)for all z D. Call F an anti-derivative of f .Theorem (Fundamental Theorem of Contour Integration)
Suppose f : D C has an anti-derivative F . Suppose is a pathfrom z0 to z1. Then
f = F (z1) F (z0).
Example
f (z) = z2. f has antiderivative F (z) = z3/3.Let be any path from 0 to i .Then
f =
F (i) F (0) = i3
3= i
3.
RemarkMost functions DO NOT have an anti-derivative. Examplef : C \ {0} C, f (z) = 1/z .
-
4 The Fundamental Theorem of Contour IntegrationDefinitionSuppose there exists a function F : D C such that F (z) = f (z)for all z D. Call F an anti-derivative of f .Theorem (Fundamental Theorem of Contour Integration)
Suppose f : D C has an anti-derivative F . Suppose is a pathfrom z0 to z1. Then
f = F (z1) F (z0).
Example
f (z) = z2. f has antiderivative F (z) = z3/3.Let be any path from 0 to i .Then
f = F (i) F (0) = i
3
3= i
3.
RemarkMost functions DO NOT have an anti-derivative.
Examplef : C \ {0} C, f (z) = 1/z .
-
4 The Fundamental Theorem of Contour IntegrationDefinitionSuppose there exists a function F : D C such that F (z) = f (z)for all z D. Call F an anti-derivative of f .Theorem (Fundamental Theorem of Contour Integration)
Suppose f : D C has an anti-derivative F . Suppose is a pathfrom z0 to z1. Then
f = F (z1) F (z0).
Example
f (z) = z2. f has antiderivative F (z) = z3/3.Let be any path from 0 to i .Then
f = F (i) F (0) = i
3
3= i
3.
RemarkMost functions DO NOT have an anti-derivative. Examplef : C \ {0} C, f (z) = 1/z .
-
4 Cauchys TheoremIf is a closed contour (= starts and ends at the same point, sayz0) AND f has an anti-derivative then by the Fund Thm ofContour Integ
f = F (z0) F (z0) = 0.
Q: What if f does NOT have an anti-derivative?
Theorem (Cauchys theorem)
Suppose:
I f : D C is holomorphicI is a closed contour in D s.t. w(, z) = 0 for all z / D.
Then
f = 0.
-
4 Cauchys TheoremIf is a closed contour (= starts and ends at the same point, sayz0) AND f has an anti-derivative then by the Fund Thm ofContour Integ
f = F (z0) F (z0) = 0.
Q: What if f does NOT have an anti-derivative?
Theorem (Cauchys theorem)
Suppose:
I f : D C is holomorphicI is a closed contour in D s.t. w(, z) = 0 for all z / D.
Then
f = 0.
-
4 Cauchys TheoremIf is a closed contour (= starts and ends at the same point, sayz0) AND f has an anti-derivative then by the Fund Thm ofContour Integ
f = F (z0) F (z0) = 0.
Q: What if f does NOT have an anti-derivative?
Theorem (Cauchys theorem)
Suppose:
I f : D C is holomorphicI is a closed contour in D s.t. w(, z) = 0 for all z / D.
Then
f = 0.
-
4 The Generalised Cauchy Theorem
Theorem (The Generalised Cauchy Theorem)
Suppose
I f : D C is holomorphicI 1, . . . , n are closed contours in D s.t.
w(1, z) + + w(n, z) = 0 for all z / D.
Then 1
f + +n
f = 0.
-
Example
1=unit circle centre 0 radius 1 describedonce anticlockwise.f (z) = 1/z on D = C \ {0}.Calculate
1
f ,2
f .
1(t) = eit , 0 t 2pi.
1f =
2pi0 f (1(t))
1(t) dt
= 2pi
01e it
ie it dt = 2pii .
Apply GCT to 1, 2 and the domain D =C \ {0}.The only point not in D is 0.w(1, 0) + w(2, 0) = 1 + 1 = 0.GCT 1 f + 2 f = 0.Hence
2
f =2
f = 2pii .
-
Example
1=unit circle centre 0 radius 1 describedonce anticlockwise.f (z) = 1/z on D = C \ {0}.Calculate
1
f ,2
f .
1(t) = eit , 0 t 2pi.
1
f = 2pi
0 f (1(t))1(t) dt
= 2pi
01e it
ie it dt = 2pii .
Apply GCT to 1, 2 and the domain D =C \ {0}.The only point not in D is 0.w(1, 0) + w(2, 0) = 1 + 1 = 0.GCT 1 f + 2 f = 0.Hence
2
f =2
f = 2pii .
-
Example
1=unit circle centre 0 radius 1 describedonce anticlockwise.f (z) = 1/z on D = C \ {0}.Calculate
1
f ,2
f .
1(t) = eit , 0 t 2pi.
1f =
2pi0 f (1(t))
1(t) dt
= 2pi
01e it
ie it dt = 2pii .
Apply GCT to 1, 2 and the domain D =C \ {0}.The only point not in D is 0.w(1, 0) + w(2, 0) = 1 + 1 = 0.GCT 1 f + 2 f = 0.Hence
2
f =2
f = 2pii .
-
Example
1=unit circle centre 0 radius 1 describedonce anticlockwise.f (z) = 1/z on D = C \ {0}.Calculate
1
f ,2
f .
1(t) = eit , 0 t 2pi.
1f =
2pi0 f (1(t))
1(t) dt
= 2pi
01e it
ie it dt = 2pii .
Apply GCT to 1, 2 and the domain D =C \ {0}.The only point not in D is 0.w(1, 0) + w(2, 0) = 1 + 1 = 0.GCT 1 f + 2 f = 0.Hence
2
f =2
f = 2pii .
-
Example
1=unit circle centre 0 radius 1 describedonce anticlockwise.f (z) = 1/z on D = C \ {0}.Calculate
1
f ,2
f .
1(t) = eit , 0 t 2pi.
1f =
2pi0 f (1(t))
1(t) dt
= 2pi
01e it
ie it dt = 2pii .
Apply GCT to 1, 2 and the domain D =C \ {0}.
The only point not in D is 0.w(1, 0) + w(2, 0) = 1 + 1 = 0.GCT 1 f + 2 f = 0.Hence
2
f =2
f = 2pii .
-
Example
1=unit circle centre 0 radius 1 describedonce anticlockwise.f (z) = 1/z on D = C \ {0}.Calculate
1
f ,2
f .
1(t) = eit , 0 t 2pi.
1f =
2pi0 f (1(t))
1(t) dt
= 2pi
01e it
ie it dt = 2pii .
Apply GCT to 1, 2 and the domain D =C \ {0}.The only point not in D is 0.
w(1, 0) + w(2, 0) = 1 + 1 = 0.GCT 1 f + 2 f = 0.Hence
2
f =2
f = 2pii .
-
Example
1=unit circle centre 0 radius 1 describedonce anticlockwise.f (z) = 1/z on D = C \ {0}.Calculate
1
f ,2
f .
1(t) = eit , 0 t 2pi.
1f =
2pi0 f (1(t))
1(t) dt
= 2pi
01e it
ie it dt = 2pii .
Apply GCT to 1, 2 and the domain D =C \ {0}.The only point not in D is 0.w(1, 0) + w(2, 0) = 1 + 1 = 0.
GCT 1 f + 2 f = 0.Hence
2
f =2
f = 2pii .
-
Example
1=unit circle centre 0 radius 1 describedonce anticlockwise.f (z) = 1/z on D = C \ {0}.Calculate
1
f ,2
f .
1(t) = eit , 0 t 2pi.
1f =
2pi0 f (1(t))
1(t) dt
= 2pi
01e it
ie it dt = 2pii .
Apply GCT to 1, 2 and the domain D =C \ {0}.The only point not in D is 0.w(1, 0) + w(2, 0) = 1 + 1 = 0.GCT 1 f + 2 f = 0.Hence
2
f =2
f = 2pii .
-
5 Cauchys Integral formula and its applications
Theorem (Cauchys Integral Formula)
Suppose
I f is holomorphic on the disc {z C | |z z0| < R}I Cr is the circular path with centre z0 radius r , r < R.
Then if w is a point s.t. |w z0| < r we have
f (w) =1
2pii
Cr
f (z)
z w dz
Proof: hard (so not examinable).
-
5 Cauchys Integral formula and its applications
Theorem (Cauchys Integral Formula)
Suppose
I f is holomorphic on the disc {z C | |z z0| < R}I Cr is the circular path with centre z0 radius r , r < R.
Then if w is a point s.t. |w z0| < r we have
f (w) =1
2pii
Cr
f (z)
z w dz
Proof: hard (so not examinable).
-
5 Cauchys Integral formula and its applications
Theorem (Cauchys Integral Formula)
Suppose
I f is holomorphic on the disc {z C | |z z0| < R}I Cr is the circular path with centre z0 radius r , r < R.
Then if w is a point s.t. |w z0| < r we have
f (w) =1
2pii
Cr
f (z)
z w dz
Proof: hard (so not examinable).
-
5 Consequences of Cauchys Integral Formula: TaylorsTheorem
Theorem (Taylors Theorem)
Suppose
I f is holomorphic on D.
Then
I all the higher derivatives of f exist in DI on any disc {z C | |z z0| < R} D, f is equal to its
Taylor series
f (z) =n=0
an(z z0)n
I
an =f (n)(z0)
n!=
1
2pii
Cr
f (z)
(z z0)n+1 dz
where Cr is a circular path centre z0 radius r < R.
Proof: hard (so not examinable).
-
5 Consequences of Cauchys Integral Formula: TaylorsTheorem
Theorem (Taylors Theorem)
Suppose
I f is holomorphic on D.
Then
I all the higher derivatives of f exist in D
I on any disc {z C | |z z0| < R} D, f is equal to itsTaylor series
f (z) =n=0
an(z z0)n
I
an =f (n)(z0)
n!=
1
2pii
Cr
f (z)
(z z0)n+1 dz
where Cr is a circular path centre z0 radius r < R.
Proof: hard (so not examinable).
-
5 Consequences of Cauchys Integral Formula: TaylorsTheorem
Theorem (Taylors Theorem)
Suppose
I f is holomorphic on D.
Then
I all the higher derivatives of f exist in DI on any disc {z C | |z z0| < R} D, f is equal to its
Taylor series
f (z) =n=0
an(z z0)n
I
an =f (n)(z0)
n!=
1
2pii
Cr
f (z)
(z z0)n+1 dz
where Cr is a circular path centre z0 radius r < R.
Proof: hard (so not examinable).
-
5 Consequences of Cauchys Integral Formula: TaylorsTheorem
Theorem (Taylors Theorem)
Suppose
I f is holomorphic on D.
Then
I all the higher derivatives of f exist in DI on any disc {z C | |z z0| < R} D, f is equal to its
Taylor series
f (z) =n=0
an(z z0)n
I
an =f (n)(z0)
n!=
1
2pii
Cr
f (z)
(z z0)n+1 dz
where Cr is a circular path centre z0 radius r < R.
Proof: hard (so not examinable).
-
5 Consequences of Cauchys Integral Formula: TaylorsTheorem
Theorem (Taylors Theorem)
Suppose
I f is holomorphic on D.
Then
I all the higher derivatives of f exist in DI on any disc {z C | |z z0| < R} D, f is equal to its
Taylor series
f (z) =n=0
an(z z0)n
I
an =f (n)(z0)
n!=
1
2pii
Cr
f (z)
(z z0)n+1 dz
where Cr is a circular path centre z0 radius r < R.
Proof: hard (so not examinable).
-
5 Consequences of Cauchys Integral Formula: CauchysEstimate
Proposition (Cauchys Estimate)
Suppose
I f is holomorphic on {z C | |z z0| < R}.I |f (z)| M for all z s.t. |z z0| = r .
Then
I for all n 0 we have
|f (n)(z0)| Mn!rn
.
Proof: easy consequence of Taylors Theorem and the EstimationLemma. (See this weeks support class sheet.)
-
5 Consequences of Cauchys Integral Formula: CauchysEstimate
Proposition (Cauchys Estimate)
Suppose
I f is holomorphic on {z C | |z z0| < R}.I |f (z)| M for all z s.t. |z z0| = r .
Then
I for all n 0 we have
|f (n)(z0)| Mn!rn
.
Proof: easy consequence of Taylors Theorem and the EstimationLemma. (See this weeks support class sheet.)
-
5 Consequences of Cauchys Integral Formula: CauchysEstimate
Proposition (Cauchys Estimate)
Suppose
I f is holomorphic on {z C | |z z0| < R}.I |f (z)| M for all z s.t. |z z0| = r .
Then
I for all n 0 we have
|f (n)(z0)| Mn!rn
.
Proof: easy consequence of Taylors Theorem and the EstimationLemma. (See this weeks support class sheet.)
-
5 Consequences of Cauchys Integral Formula: LiouvillesTheorem
Theorem (Liouvilles Theorem)
Suppose that f is holomorphic and bounded on the whole of C(i.e. M > 0 such that |f (z)| M for all z C).Then f is a constant.
Proof: easy consequence of Cauchys Estimate when n = 1.
Choose M > 0 s.t. |f (z)| M for all z C. Let z0 C.f holomorphic on whole of C implies f is holomorphic on the disc{z C | |z z0| < R} for any R > 0.By Cauchys Estimate, for any 0 < r < R we have |f (z0)| Mr .Take R, hence r , arbitrarily large. Then f (z0) = 0 z0 D.Hence f is a constant.
-
5 Consequences of Cauchys Integral Formula: LiouvillesTheorem
Theorem (Liouvilles Theorem)
Suppose that f is holomorphic and bounded on the whole of C(i.e. M > 0 such that |f (z)| M for all z C).Then f is a constant.
Proof: easy consequence of Cauchys Estimate when n = 1.
Choose M > 0 s.t. |f (z)| M for all z C. Let z0 C.f holomorphic on whole of C implies f is holomorphic on the disc{z C | |z z0| < R} for any R > 0.By Cauchys Estimate, for any 0 < r < R we have |f (z0)| Mr .Take R, hence r , arbitrarily large. Then f (z0) = 0 z0 D.Hence f is a constant.
-
5 Consequences of Cauchys Integral Formula: LiouvillesTheorem
Theorem (Liouvilles Theorem)
Suppose that f is holomorphic and bounded on the whole of C(i.e. M > 0 such that |f (z)| M for all z C).Then f is a constant.
Proof: easy consequence of Cauchys Estimate when n = 1.
Choose M > 0 s.t. |f (z)| M for all z C.
Let z0 C.f holomorphic on whole of C implies f is holomorphic on the disc{z C | |z z0| < R} for any R > 0.By Cauchys Estimate, for any 0 < r < R we have |f (z0)| Mr .Take R, hence r , arbitrarily large. Then f (z0) = 0 z0 D.Hence f is a constant.
-
5 Consequences of Cauchys Integral Formula: LiouvillesTheorem
Theorem (Liouvilles Theorem)
Suppose that f is holomorphic and bounded on the whole of C(i.e. M > 0 such that |f (z)| M for all z C).Then f is a constant.
Proof: easy consequence of Cauchys Estimate when n = 1.
Choose M > 0 s.t. |f (z)| M for all z C. Let z0 C.
f holomorphic on whole of C implies f is holomorphic on the disc{z C | |z z0| < R} for any R > 0.By Cauchys Estimate, for any 0 < r < R we have |f (z0)| Mr .Take R, hence r , arbitrarily large. Then f (z0) = 0 z0 D.Hence f is a constant.
-
5 Consequences of Cauchys Integral Formula: LiouvillesTheorem
Theorem (Liouvilles Theorem)
Suppose that f is holomorphic and bounded on the whole of C(i.e. M > 0 such that |f (z)| M for all z C).Then f is a constant.
Proof: easy consequence of Cauchys Estimate when n = 1.
Choose M > 0 s.t. |f (z)| M for all z C. Let z0 C.f holomorphic on whole of C implies f is holomorphic on the disc{z C | |z z0| < R} for any R > 0.
By Cauchys Estimate, for any 0 < r < R we have |f (z0)| Mr .Take R, hence r , arbitrarily large. Then f (z0) = 0 z0 D.Hence f is a constant.
-
5 Consequences of Cauchys Integral Formula: LiouvillesTheorem
Theorem (Liouvilles Theorem)
Suppose that f is holomorphic and bounded on the whole of C(i.e. M > 0 such that |f (z)| M for all z C).Then f is a constant.
Proof: easy consequence of Cauchys Estimate when n = 1.
Choose M > 0 s.t. |f (z)| M for all z C. Let z0 C.f holomorphic on whole of C implies f is holomorphic on the disc{z C | |z z0| < R} for any R > 0.By Cauchys Estimate, for any 0 < r < R we have |f (z0)| Mr .
Take R, hence r , arbitrarily large. Then f (z0) = 0 z0 D.Hence f is a constant.
-
5 Consequences of Cauchys Integral Formula: LiouvillesTheorem
Theorem (Liouvilles Theorem)
Suppose that f is holomorphic and bounded on the whole of C(i.e. M > 0 such that |f (z)| M for all z C).Then f is a constant.
Proof: easy consequence of Cauchys Estimate when n = 1.
Choose M > 0 s.t. |f (z)| M for all z C. Let z0 C.f holomorphic on whole of C implies f is holomorphic on the disc{z C | |z z0| < R} for any R > 0.By Cauchys Estimate, for any 0 < r < R we have |f (z0)| Mr .Take R, hence r , arbitrarily large. Then f (z0) = 0 z0 D.
Hence f is a constant.
-
5 Consequences of Cauchys Integral Formula: LiouvillesTheorem
Theorem (Liouvilles Theorem)
Suppose that f is holomorphic and bounded on the whole of C(i.e. M > 0 such that |f (z)| M for all z C).Then f is a constant.
Proof: easy consequence of Cauchys Estimate when n = 1.
Choose M > 0 s.t. |f (z)| M for all z C. Let z0 C.f holomorphic on whole of C implies f is holomorphic on the disc{z C | |z z0| < R} for any R > 0.By Cauchys Estimate, for any 0 < r < R we have |f (z0)| Mr .Take R, hence r , arbitrarily large. Then f (z0) = 0 z0 D.Hence f is a constant.
-
5 Consequences of Cauchys Integral Formula: Fund Thmof Algebra
Theorem (The Fundamental Theorem of Algebra)
Let p(z) = zn + an1zn1 + + a1z + a0 be a polynomial ofdegree n 1 with coefficients aj C.Then p has a zero: there exists w C such that p(w) = 0.
Proof: straightforward consequence of Liouvilles Theorem
Idea: Suppose not, i.e. p(z) 6= 0 for all z C. Then 1/p(z) isdefined and is holomorphic on C. Show thart 1/p(z) is bounded:there exists M > 0 such that |1/p(z)| < M for all z C. ThenLiouville implies 1/p(z) is a constant, hence p(z) is constant - acontradiction.
-
5 Consequences of Cauchys Integral Formula: Fund Thmof Algebra
Theorem (The Fundamental Theorem of Algebra)
Let p(z) = zn + an1zn1 + + a1z + a0 be a polynomial ofdegree n 1 with coefficients aj C.Then p has a zero: there exists w C such that p(w) = 0.
Proof: straightforward consequence of Liouvilles Theorem
Idea: Suppose not, i.e. p(z) 6= 0 for all z C. Then 1/p(z) isdefined and is holomorphic on C. Show thart 1/p(z) is bounded:there exists M > 0 such that |1/p(z)| < M for all z C. ThenLiouville implies 1/p(z) is a constant, hence p(z) is constant - acontradiction.
-
5 Consequences of Cauchys Integral Formula: Fund Thmof Algebra
Theorem (The Fundamental Theorem of Algebra)
Let p(z) = zn + an1zn1 + + a1z + a0 be a polynomial ofdegree n 1 with coefficients aj C.Then p has a zero: there exists w C such that p(w) = 0.
Proof: straightforward consequence of Liouvilles Theorem
Idea: Suppose not, i.e. p(z) 6= 0 for all z C. Then 1/p(z) isdefined and is holomorphic on C. Show thart 1/p(z) is bounded:there exists M > 0 such that |1/p(z)| < M for all z C. ThenLiouville implies 1/p(z) is a constant, hence p(z) is constant - acontradiction.
-
6 Laurent SeriesA power series at z0 has the form
n=0 an(z z0)n.
This converges for {z C | |z z0| < R}.
A Laurent series has the form
n=an(z z0)n.
Often we write this instead as
n=1
bn(z z0)n +
n=0
an(z z0)n
= + bn(z z0)n + +
b1z z0 +a0+a1(zz0)+ +an(zz0)
n+ .
A Laurent series converges on an annulus:{z C | R1 < |z z0| < R2}.(A punctured disc {z C | 0 < |z z0| < R2} is an annulus.)
-
6 Laurent SeriesA power series at z0 has the form
n=0 an(z z0)n.
This converges for {z C | |z z0| < R}.
A Laurent series has the form
n=an(z z0)n.
Often we write this instead as
n=1
bn(z z0)n +
n=0
an(z z0)n
= + bn(z z0)n + +
b1z z0 +a0+a1(zz0)+ +an(zz0)
n+ .
A Laurent series converges on an annulus:{z C | R1 < |z z0| < R2}.(A punctured disc {z C | 0 < |z z0| < R2} is an annulus.)
-
6 Laurent SeriesA power series at z0 has the form
n=0 an(z z0)n.
This converges for {z C | |z z0| < R}.
A Laurent series has the form
n=an(z z0)n.
Often we write this instead as
n=1
bn(z z0)n +
n=0
an(z z0)n
= + bn(z z0)n + +
b1z z0 +a0+a1(zz0)+ +an(zz0)
n+ .
A Laurent series converges on an annulus:{z C | R1 < |z z0| < R2}.(A punctured disc {z C | 0 < |z z0| < R2} is an annulus.)
-
6 Laurent SeriesA power series at z0 has the form
n=0 an(z z0)n.
This converges for {z C | |z z0| < R}.
A Laurent series has the form
n=an(z z0)n.
Often we write this instead as
n=1
bn(z z0)n +
n=0
an(z z0)n
= + bn(z z0)n + +
b1z z0 +a0+a1(zz0)+ +an(zz0)
n+ .
A Laurent series converges on an annulus:{z C | R1 < |z z0| < R2}.(A punctured disc {z C | 0 < |z z0| < R2} is an annulus.)
-
6 Laurent SeriesA power series at z0 has the form
n=0 an(z z0)n.
This converges for {z C | |z z0| < R}.
A Laurent series has the form
n=an(z z0)n.
Often we write this instead as
n=1
bn(z z0)n +
n=0
an(z z0)n
= + bn(z z0)n + +
b1z z0 +a0+a1(zz0)+ +an(zz0)
n+ .
A Laurent series converges on an annulus:{z C | R1 < |z z0| < R2}.(A punctured disc {z C | 0 < |z z0| < R2} is an annulus.)
-
6 Laurent SeriesA power series at z0 has the form
n=0 an(z z0)n.
This converges for {z C | |z z0| < R}.
A Laurent series has the form
n=an(z z0)n.
Often we write this instead as
n=1
bn(z z0)n +
n=0
an(z z0)n
= + bn(z z0)n + +
b1z z0 +a0+a1(zz0)+ +an(zz0)
n+ .
A Laurent series converges on an annulus:{z C | R1 < |z z0| < R2}.
(A punctured disc {z C | 0 < |z z0| < R2} is an annulus.)
-
6 Laurent SeriesA power series at z0 has the form
n=0 an(z z0)n.
This converges for {z C | |z z0| < R}.
A Laurent series has the form
n=an(z z0)n.
Often we write this instead as
n=1
bn(z z0)n +
n=0
an(z z0)n
= + bn(z z0)n + +
b1z z0 +a0+a1(zz0)+ +an(zz0)
n+ .
A Laurent series converges on an annulus:{z C | R1 < |z z0| < R2}.(A punctured disc {z C | 0 < |z z0| < R2} is an annulus.)
-
6 Laurents Theorem
Theorem (Laurents Theorem)
Suppose f is holomorphic on the annulus{z C | R1 < |z z0| < R2}, 0 R1 < R2 .
Then we can write f as a Laurent series
f (z) =n=1
an(z z0)n +n=0
an(z z0)n
on {z C | R1 < |z z0| < R2}.Moreover
an =1
2pii
Cr
f (z)
(z z0)n+1 dz
where Cr is the circular path centre z0 radius r (R1 < r < R2)described once anticlockwise.
-
6 Laurents Theorem
Theorem (Laurents Theorem)
Suppose f is holomorphic on the annulus{z C | R1 < |z z0| < R2}, 0 R1 < R2 .Then we can write f as a Laurent series
f (z) =n=1
an(z z0)n +n=0
an(z z0)n
on {z C | R1 < |z z0| < R2}.
Moreover
an =1
2pii
Cr
f (z)
(z z0)n+1 dz
where Cr is the circular path centre z0 radius r (R1 < r < R2)described once anticlockwise.
-
6 Laurents Theorem
Theorem (Laurents Theorem)
Suppose f is holomorphic on the annulus{z C | R1 < |z z0| < R2}, 0 R1 < R2 .Then we can write f as a Laurent series
f (z) =n=1
an(z z0)n +n=0
an(z z0)n
on {z C | R1 < |z z0| < R2}.Moreover
an =1
2pii
Cr
f (z)
(z z0)n+1 dz
where Cr is the circular path centre z0 radius r (R1 < r < R2)described once anticlockwise.
-
6 Laurents Theorem: an example
Let f (z) = 1z(1z) .
This is holomorphic on 0 < |z | < 1.Note
1
1 z = 1 + z + z2 + z3 + if |z | < 1.
So
f (z) =1
z(1 + z + z2 + z3 + ) = 1
z+ 1 + z + z2 + z3 +
if 0 < |z | < 1. This is the Laurent series on the annulus0 < |z | < 1.
-
6 Laurents Theorem: an example
Let f (z) = 1z(1z) .
This is holomorphic on 0 < |z | < 1.
Note1
1 z = 1 + z + z2 + z3 + if |z | < 1.
So
f (z) =1
z(1 + z + z2 + z3 + ) = 1
z+ 1 + z + z2 + z3 +
if 0 < |z | < 1. This is the Laurent series on the annulus0 < |z | < 1.
-
6 Laurents Theorem: an example
Let f (z) = 1z(1z) .
This is holomorphic on 0 < |z | < 1.Note
1
1 z = 1 + z + z2 + z3 + if |z | < 1.
So
f (z) =1
z(1 + z + z2 + z3 + ) = 1
z+ 1 + z + z2 + z3 +
if 0 < |z | < 1. This is the Laurent series on the annulus0 < |z | < 1.
-
6 Laurents Theorem: an example
Let f (z) = 1z(1z) .
This is holomorphic on 0 < |z | < 1.Note
1
1 z = 1 + z + z2 + z3 + if |z | < 1.
So
f (z) =1
z(1 + z + z2 + z3 + ) = 1
z+ 1 + z + z2 + z3 +
if 0 < |z | < 1.
This is the Laurent series on the annulus0 < |z | < 1.
-
6 Laurents Theorem: an example
Let f (z) = 1z(1z) .
This is holomorphic on 0 < |z | < 1.Note
1
1 z = 1 + z + z2 + z3 + if |z | < 1.
So
f (z) =1
z(1 + z + z2 + z3 + ) = 1
z+ 1 + z + z2 + z3 +
if 0 < |z | < 1. This is the Laurent series on the annulus0 < |z | < 1.
-
6 Laurents Theorem: an example
Let f (z) = 1z(1z) .
f is also holomorphic on |z | > 1.Note if |1/z | < 1 (i.e. |z | > 1)
1
1 z =1
z
(1
1 1z
)=
1
z
(1 +
1
z+
1
z2+
1
z3+
).
=1
z+
1
z2+
1
z3+
1
z4+
Hence
f (z) =1
z2+
1
z3+
1
z4+
1
z5+
if 1 < |z |
-
6 Laurents Theorem: an example
Let f (z) = 1z(1z) .
f is also holomorphic on |z | > 1.
Note if |1/z | < 1 (i.e. |z | > 1)
1
1 z =1
z
(1
1 1z
)=
1
z
(1 +
1
z+
1
z2+
1
z3+
).
=1
z+
1
z2+
1
z3+
1
z4+
Hence
f (z) =1
z2+
1
z3+
1
z4+
1
z5+
if 1 < |z |
-
6 Laurents Theorem: an example
Let f (z) = 1z(1z) .
f is also holomorphic on |z | > 1.Note if |1/z | < 1 (i.e. |z | > 1)
1
1 z =1
z
(1
1 1z
)=
1
z
(1 +
1
z+
1
z2+
1
z3+
).
=1
z+
1
z2+
1
z3+
1
z4+
Hence
f (z) =1
z2+
1
z3+
1
z4+
1
z5+
if 1 < |z |
-
6 Laurents Theorem: an example
Let f (z) = 1z(1z) .
f is also holomorphic on |z | > 1.Note if |1/z | < 1 (i.e. |z | > 1)
1
1 z =1
z
(1
1 1z
)=
1
z
(1 +
1
z+
1
z2+
1
z3+
).
=1
z+
1
z2+
1
z3+
1
z4+
Hence
f (z) =1
z2+
1
z3+
1
z4+
1
z5+
if 1 < |z |
-
6 Laurents Theorem: an example
Let f (z) = 1z(1z) .
f is also holomorphic on |z | > 1.Note if |1/z | < 1 (i.e. |z | > 1)
1
1 z =1
z
(1
1 1z
)=
1
z
(1 +
1
z+
1
z2+
1
z3+
).
=1
z+
1
z2+
1
z3+
1
z4+
Hence
f (z) =1
z2+
1
z3+
1
z4+
1
z5+
if 1 < |z |
-
7 Poles and residuesDefinitionf has a singularity at z0 if f is not differentiable at z0.
RemarkFor us, singularities normally occur when f is not even defined atz0 (usually because we are dividing by 0).
z0 is an isolated singularity if there are no other singularitiesnearby: i.e. there exists R > 0 such that f is holomorphic on thepunctured disc 0 < |z z0| < R.Apply Laurents Theorem to this punctured disc:
f (z) =n=1
bn(z z0)n +
n=0
an(z z0)n
= principal part + a power series
If the principal part has finitely many terms in it then z0 is a pole.
-
7 Poles and residuesDefinitionf has a singularity at z0 if f is not differentiable at z0.
RemarkFor us, singularities normally occur when f is not even defined atz0 (usually because we are dividing by 0).
z0 is an isolated singularity if there are no other singularitiesnearby: i.e. there exists R > 0 such that f is holomorphic on thepunctured disc 0 < |z z0| < R.Apply Laurents Theorem to this punctured disc:
f (z) =n=1
bn(z z0)n +
n=0
an(z z0)n
= principal part + a power series
If the principal part has finitely many terms in it then z0 is a pole.
-
7 Poles and residuesDefinitionf has a singularity at z0 if f is not differentiable at z0.
RemarkFor us, singularities normally occur when f is not even defined atz0 (usually because we are dividing by 0).
z0 is an isolated singularity if there are no other singularitiesnearby:
i.e. there exists R > 0 such that f is holomorphic on thepunctured disc 0 < |z z0| < R.Apply Laurents Theorem to this punctured disc:
f (z) =n=1
bn(z z0)n +
n=0
an(z z0)n
= principal part + a power series
If the principal part has finitely many terms in it then z0 is a pole.
-
7 Poles and residuesDefinitionf has a singularity at z0 if f is not differentiable at z0.
RemarkFor us, singularities normally occur when f is not even defined atz0 (usually because we are dividing by 0).
z0 is an isolated singularity if there are no other singularitiesnearby: i.e. there exists R > 0 such that f is holomorphic on thepunctured disc 0 < |z z0| < R.
Apply Laurents Theorem to this punctured disc:
f (z) =n=1
bn(z z0)n +
n=0
an(z z0)n
= principal part + a power series
If the principal part has finitely many terms in it then z0 is a pole.
-
7 Poles and residuesDefinitionf has a singularity at z0 if f is not differentiable at z0.
RemarkFor us, singularities normally occur when f is not even defined atz0 (usually because we are dividing by 0).
z0 is an isolated singularity if there are no other singularitiesnearby: i.e. there exists R > 0 such that f is holomorphic on thepunctured disc 0 < |z z0| < R.Apply Laurents Theorem to this punctured disc:
f (z) =n=1
bn(z z0)n +
n=0
an(z z0)n
= principal part + a power series
If the principal part has finitely many terms in it then z0 is a pole.
-
7 Poles and residuesDefinitionf has a singularity at z0 if f is not differentiable at z0.
RemarkFor us, singularities normally occur when f is not even defined atz0 (usually because we are dividing by 0).
z0 is an isolated singularity if there are no other singularitiesnearby: i.e. there exists R > 0 such that f is holomorphic on thepunctured disc 0 < |z z0| < R.Apply Laurents Theorem to this punctured disc:
f (z) =n=1
bn(z z0)n +
n=0
an(z z0)n
= principal part + a power series
If the principal part has finitely many terms in it then z0 is a pole.
-
7 Poles and residues
Suppose f has a pole of order m at z0. Then on 0 < |z z0| < Rfor some R > 0 we can write
f (z) =bm
(z z0)m + +b1
z z0 +a0+a1(zz0)+ +an(zz0)n+
where bm 6= 0.
m=order of the pole (most negative power).
b1 = residue of the pole (coefficient of (z z0)1 term) =Res(f , z0).
-
7 Poles and residues
Suppose f has a pole of order m at z0. Then on 0 < |z z0| < Rfor some R > 0 we can write
f (z) =bm
(z z0)m + +b1
z z0 +a0+a1(zz0)+ +an(zz0)n+
where bm 6= 0.m=order of the pole (most negative power).
b1 = residue of the pole (coefficient of (z z0)1 term) =Res(f , z0).
-
7 Poles and residues
Suppose f has a pole of order m at z0. Then on 0 < |z z0| < Rfor some R > 0 we can write
f (z) =bm
(z z0)m + +b1
z z0 +a0+a1(zz0)+ +an(zz0)n+
where bm 6= 0.m=order of the pole (most negative power).
b1 = residue of the pole (coefficient of (z z0)1 term) =Res(f , z0).
-
7 Poles and residues and zeroesq(z) has a zero of order m at z0 if we can write it asq(z) = (z z0)mg(z) where g(z0) 6= 0.
Example: q(z) = (z2 + 25)2 has a zero of order 2 at 5i . (Why?(z2 + 25)2 = (z + 5i)2(z 5i)2.)
LemmaSuppose f (z) = p(z)/q(z) where p, q holomorphic. Suppose
I q has a zero of order m at z0I p(z0) 6= 0.
Then f has a pole of order m at z0.
Example:
f (z) =z + 1
(z2 + 25)2
has poles of order 2 at 5i .
-
7 Poles and residues and zeroesq(z) has a zero of order m at z0 if we can write it asq(z) = (z z0)mg(z) where g(z0) 6= 0.Example: q(z) = (z2 + 25)2 has a zero of order 2 at 5i .
(Why?(z2 + 25)2 = (z + 5i)2(z 5i)2.)
LemmaSuppose f (z) = p(z)/q(z) where p, q holomorphic. Suppose
I q has a zero of order m at z0I p(z0) 6= 0.
Then f has a pole of order m at z0.
Example:
f (z) =z + 1
(z2 + 25)2
has poles of order 2 at 5i .
-
7 Poles and residues and zeroesq(z) has a zero of order m at z0 if we can write it asq(z) = (z z0)mg(z) where g(z0) 6= 0.Example: q(z) = (z2 + 25)2 has a zero of order 2 at 5i . (Why?(z2 + 25)2 = (z + 5i)2(z 5i)2.)
LemmaSuppose f (z) = p(z)/q(z) where p, q holomorphic. Suppose
I q has a zero of order m at z0I p(z0) 6= 0.
Then f has a pole of order m at z0.
Example:
f (z) =z + 1
(z2 + 25)2
has poles of order 2 at 5i .
-
7 Poles and residues and zeroesq(z) has a zero of order m at z0 if we can write it asq(z) = (z z0)mg(z) where g(z0) 6= 0.Example: q(z) = (z2 + 25)2 has a zero of order 2 at 5i . (Why?(z2 + 25)2 = (z + 5i)2(z 5i)2.)
LemmaSuppose f (z) = p(z)/q(z) where p, q holomorphic. Suppose
I q has a zero of order m at z0I p(z0) 6= 0.
Then f has a pole of order m at z0.
Example:
f (z) =z + 1
(z2 + 25)2
has poles of order 2 at 5i .
-
7 Poles and residues and zeroesq(z) has a zero of order m at z0 if we can write it asq(z) = (z z0)mg(z) where g(z0) 6= 0.Example: q(z) = (z2 + 25)2 has a zero of order 2 at 5i . (Why?(z2 + 25)2 = (z + 5i)2(z 5i)2.)
LemmaSuppose f (z) = p(z)/q(z) where p, q holomorphic. Suppose
I q has a zero of order m at z0I p(z0) 6= 0.
Then f has a pole of order m at z0.
Example:
f (z) =z + 1
(z2 + 25)2
has poles of order 2 at 5i .
-
7 Calculating residues
We had various formulae for calculating residues:
LemmaIf f has a simple pole at z0 then Res(f , z0) = limzz0(z z0)f (z).
If f (z) has a pole of order m at z0 then
Res(f , z0) = limzz0
(1
(m 1)!dm1
dzm1((z z0)mf (z))
).
-
7 Calculating residues
We had various formulae for calculating residues:
LemmaIf f has a simple pole at z0 then Res(f , z0) = limzz0(z z0)f (z).
If f (z) has a pole of order m at z0 then
Res(f , z0) = limzz0
(1
(m 1)!dm1
dzm1((z z0)mf (z))
).
-
7 Calculating residues
We had various formulae for calculating residues:
LemmaIf f has a simple pole at z0 then Res(f , z0) = limzz0(z z0)f (z).
If f (z) has a pole of order m at z0 then
Res(f , z0) = limzz0
(1
(m 1)!dm1
dzm1((z z0)mf (z))
).
-
7 Calculating residuesSuppose f has a pole of order 2 at z0.
Then f has a Laurent series
f (z) =b2
(z z0)2 +b1
z z0 + a0 + a1(z z0) + a2(z z0)2 +
on 0 < |z z0| < R.
(zz0)2f (z) = b2+b1(zz0)+a0(zz0)2+a1(zz0)3+a2(zz0)4+ .
d
dz
((z z0)2f (z)
)= b1 + 2a0(z z0) + 3a1(z z0)2 +
limzz0
d
dz
((z z0)2f (z)
)= b1.
-
7 Calculating residuesSuppose f has a pole of order 2 at z0. Then f has a Laurent series
f (z) =b2
(z z0)2 +b1
z z0 + a0 + a1(z z0) + a2(z z0)2 +
on 0 < |z z0| < R.
(zz0)2f (z) = b2+b1(zz0)+a0(zz0)2+a1(zz0)3+a2(zz0)4+ .
d
dz
((z z0)2f (z)
)= b1 + 2a0(z z0) + 3a1(z z0)2 +
limzz0
d
dz
((z z0)2f (z)
)= b1.
-
7 Calculating residuesSuppose f has a pole of order 2 at z0. Then f has a Laurent series
f (z) =b2
(z z0)2 +b1
z z0 + a0 + a1(z z0) + a2(z z0)2 +
on 0 < |z z0| < R.
(zz0)2f (z) = b2+b1(zz0)+a0(zz0)2+a1(zz0)3+a2(zz0)4+ .
d
dz
((z z0)2f (z)
)= b1 + 2a0(z z0) + 3a1(z z0)2 +
limzz0
d
dz
((z z0)2f (z)
)= b1.
-
7 Calculating residuesSuppose f has a pole of order 2 at z0. Then f has a Laurent series
f (z) =b2
(z z0)2 +b1
z z0 + a0 + a1(z z0) + a2(z z0)2 +
on 0 < |z z0| < R.
(zz0)2f (z) = b2+b1(zz0)+a0(zz0)2+a1(zz0)3+a2(zz0)4+ .
d
dz
((z z0)2f (z)
)= b1 + 2a0(z z0) + 3a1(z z0)2 +
limzz0
d
dz
((z z0)2f (z)
)= b1.
-
7 Calculating residuesSuppose f has a pole of order 2 at z0. Then f has a Laurent series
f (z) =b2
(z z0)2 +b1
z z0 + a0 + a1(z z0) + a2(z z0)2 +
on 0 < |z z0| < R.
(zz0)2f (z) = b2+b1(zz0)+a0(zz0)2+a1(zz0)3+a2(zz0)4+ .
d
dz
((z z0)2f (z)
)= b1 + 2a0(z z0) + 3a1(z z0)2 +
limzz0
d
dz
((z z0)2f (z)
)= b1.
-
7 Cauchys Residue Theorem
Theorem (Cauchys Residue Theorem)
Let D be a domain. Suppose
I f : D C is meromorphic (=holomorphic except for finitelymany removable singularities or poles)
I is a simple closed loop in D s.t. all the points inside lie inD
I f has poles at z1, . . . , zn inside .
Then
f (z) dz = 2piin
j=1
Res(f , zj).
-
7 Cauchys Residue Theorem
Theorem (Cauchys Residue Theorem)
Let D be a domain. Suppose
I f : D C is meromorphic (=holomorphic except for finitelymany removable singularities or poles)
I is a simple closed loop in D s.t. all the points inside lie inD
I f has poles at z1, . . . , zn inside .
Then
f (z) dz = 2piin
j=1
Res(f , zj).
-
7 Cauchys Residue Theorem
Theorem (Cauchys Residue Theorem)
Let D be a domain. Suppose
I f : D C is meromorphic (=holomorphic except for finitelymany removable singularities or poles)
I is a simple closed loop in D s.t. all the points inside lie inD
I f has poles at z1, . . . , zn inside .
Then
f (z) dz = 2piin
j=1
Res(f , zj).
-
7 Cauchys Residue Theorem
Theorem (Cauchys Residue Theorem)
Let D be a domain. Suppose
I f : D C is meromorphic (=holomorphic except for finitelymany removable singularities or poles)
I is a simple closed loop in D s.t. all the points inside lie inD
I f has poles at z1, . . . , zn inside .
Then
f (z) dz = 2piin
j=1
Res(f , zj).
-
7 Cauchys Residue Theorem
Theorem (Cauchys Residue Theorem)
Let D be a domain. Suppose
I f : D C is meromorphic (=holomorphic except for finitelymany removable singularities or poles)
I is a simple closed loop in D s.t. all the points inside lie inD
I f has poles at z1, . . . , zn inside .
Then
f (z) dz = 2piin
j=1
Res(f , zj).
-
7 Cauchys Residue Theorem: an easy exampleTake f (z) = e
iz
(z3)2 .
Then f has a pole of order 2 at z = 3.
Res(f , 3) = limz3
d
dz(z 3)2f (z) = lim
z3d
dze iz = lim
z3ie iz = ie3i .
Let C5 be the circle centre 0 radius 5, described onceanticlockwise. Then the pole at z = 3 lies inside . So by the CRT
C5
f = 2pii Res(f , 3) = 2pie3i .
Let C2 be the circle centre 0 radius 2, described once anticlockwise.Then there are no poles of f inside C2. So by the CRT
C2
f = 0.
-
7 Cauchys Residue Theorem: an easy exampleTake f (z) = e
iz
(z3)2 .
Then f has a pole of order 2 at z = 3.
Res(f , 3) = limz3
d
dz(z 3)2f (z) = lim
z3d
dze iz = lim
z3ie iz = ie3i .
Let C5 be the circle centre 0 radius 5, described onceanticlockwise. Then the pole at z = 3 lies inside . So by the CRT
C5
f = 2pii Res(f , 3) = 2pie3i .
Let C2 be the circle centre 0 radius 2, described once anticlockwise.Then there are no poles of f inside C2. So by the CRT
C2
f = 0.
-
7 Cauchys Residue Theorem: an easy exampleTake f (z) = e
iz
(z3)2 .
Then f has a pole of order 2 at z = 3.
Res(f , 3) = limz3
d
dz(z 3)2f (z)
= limz3
d
dze iz = lim
z3ie iz = ie3i .
Let C5 be the circle centre 0 radius 5, described onceanticlockwise. Then the pole at z = 3 lies inside . So by the CRT
C5
f = 2pii Res(f , 3) = 2pie3i .
Let C2 be the circle centre 0 radius 2, described once anticlockwise.Then there are no poles of f inside C2. So by the CRT
C2
f = 0.
-
7 Cauchys Residue Theorem: an easy exampleTake f (z) = e
iz
(z3)2 .
Then f has a pole of order 2 at z = 3.
Res(f , 3) = limz3
d
dz(z 3)2f (z) = lim
z3d
dze iz = lim
z3ie iz = ie3i .
Let C5 be the circle centre 0 radius 5, described onceanticlockwise. Then the pole at z = 3 lies inside . So by the CRT
C5
f = 2pii Res(f , 3) = 2pie3i .
Let C2 be the circle centre 0 radius 2, described once anticlockwise.Then there are no poles of f inside C2. So by the CRT
C2
f = 0.
-
7 Cauchys Residue Theorem: an easy exampleTake f (z) = e
iz
(z3)2 .
Then f has a pole of order 2 at z = 3.
Res(f , 3) = limz3
d
dz(z 3)2f (z) = lim
z3d
dze iz = lim
z3ie iz = ie3i .
Let C5 be the circle centre 0 radius 5, described onceanticlockwise.
Then the pole at z = 3 lies inside . So by the CRTC5
f = 2pii Res(f , 3) = 2pie3i .
Let C2 be the circle centre 0 radius 2, described once anticlockwise.Then there are no poles of f inside C2. So by the CRT
C2
f = 0.
-
7 Cauchys Residue Theorem: an easy exampleTake f (z) = e
iz
(z3)2 .
Then f has a pole of order 2 at z = 3.
Res(f , 3) = limz3
d
dz(z 3)2f (z) = lim
z3d
dze iz = lim
z3ie iz = ie3i .
Let C5 be the circle centre 0 radius 5, described onceanticlockwise. Then the pole at z = 3 lies inside . So by the CRT
C5
f = 2pii Res(f , 3) = 2pie3i .
Let C2 be the circle centre 0 radius 2, described once anticlockwise.Then there are no poles of f inside C2. So by the CRT
C2
f = 0.
-
7 Cauchys Residue Theorem: an easy exampleTake f (z) = e
iz
(z3)2 .
Then f has a pole of order 2 at z = 3.
Res(f , 3) = limz3
d
dz(z 3)2f (z) = lim
z3d
dze iz = lim
z3ie iz = ie3i .
Let C5 be the circle centre 0 radius 5, described onceanticlockwise. Then the pole at z = 3 lies inside . So by the CRT
C5
f = 2pii Res(f , 3) = 2pie3i .
Let C2 be the circle centre 0 radius 2, described once anticlockwise.
Then there are no poles of f inside C2. So by the CRTC2
f = 0.
-
7 Cauchys Residue Theorem: an easy exampleTake f (z) = e
iz
(z3)2 .
Then f has a pole of order 2 at z = 3.
Res(f , 3) = limz3
d
dz(z 3)2f (z) = lim
z3d
dze iz = lim
z3ie iz = ie3i .
Let C5 be the circle centre 0 radius 5, described onceanticlockwise. Then the pole at z = 3 lies inside . So by the CRT
C5
f = 2pii Res(f , 3) = 2pie3i .
Let C2 be the circle centre 0 radius 2, described once anticlockwise.Then there are no poles of f inside C2. So by the CRT
C2
f = 0.
-
7 Applications of the CRT: calculating infinite integralsRecall
f (x) dx = lim
A,B
BA
f (x) dx (1)
where the limit is taken in either order.
The principal value of this integral is defined to be
limR
RR
f (x) dx (2)
(2) may exist even if (1) doesnt (eg: f (x) = x).
Lemma (Technical Lemma)
Suppose f : R C is such that: there exist K > 0, C > 0, r > 1s.t. for all |x | > K we have |f (x)| < C/|x |r . Then (1) exists and isequal to its principal value.
-
7 Applications of the CRT: calculating infinite integralsRecall
f (x) dx = lim
A,B
BA
f (x) dx (1)
where the limit is taken in either order.
The principal value of this integral is defined to be
limR
RR
f (x) dx (2)
(2) may exist even if (1) doesnt (eg: f (x) = x).
Lemma (Technical Lemma)
Suppose f : R C is such that: there exist K > 0, C > 0, r > 1s.t. for all |x | > K we have |f (x)| < C/|x |r . Then (1) exists and isequal to its principal value.
-
7 Applications of the CRT: calculating infinite integralsRecall
f (x) dx = lim
A,B
BA
f (x) dx (1)
where the limit is taken in either order.
The principal value of this integral is defined to be
limR
RR
f (x) dx (2)
(2) may exist even if (1) doesnt (eg: f (x) = x).
Lemma (Technical Lemma)
Suppose f : R C is such that: there exist K > 0, C > 0, r > 1s.t. for all |x | > K we have |f (x)| < C/|x |r . Then (1) exists and isequal to its principal value.
-
7 Applications of the CRT: calculating infinite integralsRecall
f (x) dx = lim
A,B
BA
f (x) dx (1)
where the limit is taken in either order.
The principal value of this integral is defined to be
limR
RR
f (x) dx (2)
(2) may exist even if (1) doesnt (eg: f (x) = x).
Lemma (Technical Lemma)
Suppose f : R C is such that: there exist K > 0, C > 0, r > 1s.t. for all |x | > K we have |f (x)| < C/|x |r . Then (1) exists and isequal to its principal value.
-
7 Applications of the CRT: calculating infinite integralsTo calculate
f (x) dx :
(i) Check that f satisfies the technical lemma.
(ii) Construct a D-shaped contour R = [R,R] + SR .(iii) Find the poles and residues of f inside R for large R.
(iv) Use the CRT to calculate
Rf .
(v) Write R
f =
[R,R]
f +
SR
f .
(vi) Show (using the Estimation lemma) thatSR
f 0 asR .
(vii) Hence calculate
f (x) dx = limR
[R,R]
f .
See Exercises 7.4, 7.5, 7.6, 7.7.
-
7 Applications of the CRT: calculating infinite integralsTo calculate
f (x) dx :
(i) Check that f satisfies the technical lemma.
(ii) Construct a D-shaped contour R = [R,R] + SR .(iii) Find the poles and residues of f inside R for large R.
(iv) Use the CRT to calculate
Rf .
(v) Write R
f =
[R,R]
f +
SR
f .
(vi) Show (using the Estimation lemma) thatSR
f 0 asR .
(vii) Hence calculate
f (x) dx = limR
[R,R]
f .
See Exercises 7.4, 7.5, 7.6, 7.7.
-
7 Applications of the CRT: calculating infinite integralsTo calculate
f (x) dx :
(i) Check that f satisfies the technical lemma.
(ii) Construct a D-shaped contour R = [R,R] + SR .
(iii) Find the poles and residues of f inside R for large R.
(iv) Use the CRT to calculate
Rf .
(v) Write R
f =
[R,R]
f +
SR
f .
(vi) Show (using the Estimation lemma) thatSR
f 0 asR .
(vii) Hence calculate
f (x) dx = limR
[R,R]
f .
See Exercises 7.4, 7.5, 7.6, 7.7.
-
7 Applications of the CRT: calculating infinite integralsTo calculate
f (x) dx :
(i) Check that f satisfies the technical lemma.
(ii) Construct a D-shaped contour R = [R,R] + SR .(iii) Find the poles and residues of f inside R for large R.
(iv) Use the CRT to calculate
Rf .
(v) Write R
f =
[R,R]
f +
SR
f .
(vi) Show (using the Estimation lemma) thatSR
f 0 asR .
(vii) Hence calculate
f (x) dx = limR
[R,R]
f .
See Exercises 7.4, 7.5, 7.6, 7.7.
-
7 Applications of the CRT: calculating infinite integralsTo calculate
f (x) dx :
(i) Check that f satisfies the technical lemma.
(ii) Construct a D-shaped contour R = [R,R] + SR .(iii) Find the poles and residues of f inside R for large R.
(iv) Use the CRT to calculate
Rf .
(v) Write R
f =
[R,R]
f +
SR
f .
(vi) Show (using the Estimation lemma) thatSR
f 0 asR .
(vii) Hence calculate
f (x) dx = limR
[R,R]
f .
See Exercises 7.4, 7.5, 7.6, 7.7.
-
7 Applications of the CRT: calculating infinite integralsTo calculate
f (x) dx :
(i) Check that f satisfies the technical lemma.
(ii) Construct a D-shaped contour R = [R,R] + SR .(iii) Find the poles and residues of f inside R for large R.
(iv) Use the CRT to calculate
Rf .
(v) Write R
f =
[R,R]
f +
SR
f .
(vi) Show (using the Estimation lemma) thatSR
f 0 asR .
(vii) Hence calculate
f (x) dx = limR
[R,R]
f .
See Exercises 7.4, 7.5, 7.6, 7.7.
-
7 Applications of the CRT: calculating infinite integralsTo calculate
f (x) dx :
(i) Check that f satisfies the technical lemma.
(ii) Construct a D-shaped contour R = [R,R] + SR .(iii) Find the poles and residues of f inside R for large R.
(iv) Use the CRT to calculate
Rf .
(v) Write R
f =
[R,R]
f +
SR
f .
(vi) Show (using the Estimation lemma) thatSR
f 0 asR .
(vii) Hence calculate
f (x) dx = limR
[R,R]
f .
See Exercises 7.4, 7.5, 7.6, 7.7.
-
7 Applications of the CRT: calculating infinite integralsTo calculate
f (x) dx :
(i) Check that f satisfies the technical lemma.
(ii) Construct a D-shaped contour R = [R,R] + SR .(iii) Find the poles and residues of f inside R for large R.
(iv) Use the CRT to calculate
Rf .
(v) Write R
f =
[R,R]
f +
SR
f .
(vi) Show (using the Estimation lemma) thatSR
f 0 asR .
(vii) Hence calculate
f (x) dx = limR
[R,R]
f .
See Exercises 7.4, 7.5, 7.6, 7.7.
-
7 Applications of the CRT: calculating infinite integralsTo calculate
f (x) dx :
(i) Check that f satisfies the technical lemma.
(ii) Construct a D-shaped contour R = [R,R] + SR .(iii) Find the poles and residues of f inside R for large R.
(iv) Use the CRT to calculate
Rf .
(v) Write R
f =
[R,R]
f +
SR
f .
(vi) Show (using the Estimation lemma) thatSR
f 0 asR .
(vii) Hence calculate
f (x) dx = limR
[R,R]
f .
See Exercises 7.4, 7.5, 7.6, 7.7.
-
Which proofs are examinable?
Proofs of all theorems, propositions, lemmas, corollaries, etc, that Idid in the lectures are examinable EXCEPT FOR:
I Proof of Theorem 3.3.2I Proof of Lemma 4.4.1I Proof of Theorem 5.1.1I Proof of Theorem 5.2.1
ALL OTHER PROOFS THAT I DID IN THE LECTURESARE EXAMINABLE. ANYTHING I DID IN THELECTURES IS EXAMINABLE. YOU SHOULD ALSOMAKE SURE YOUVE DONE THE EXERCISES. (SODONT COMPLAIN I HAVENT WARNED YOU WHAT ISAND WHAT ISNT EXAMINABLE!)
-
Which proofs are examinable?
Proofs of all theorems, propositions, lemmas, corollaries, etc, that Idid in the lectures are examinable EXCEPT FOR:
I Proof of Theorem 3.3.2I Proof of Lemma 4.4.1I Proof of Theorem 5.1.1I Proof of Theorem 5.2.1
ALL OTHER PROOFS THAT I DID IN THE LECTURESARE EXAMINABLE. ANYTHING I DID IN THELECTURES IS EXAMINABLE. YOU SHOULD ALSOMAKE SURE YOUVE DONE THE EXERCISES. (SODONT COMPLAIN I HAVENT WARNED YOU WHAT ISAND WHAT ISNT EXAMINABLE!)