Lecture 23 25
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Transcript of Lecture 23 25
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MATH F112 (Mathematics-II)
Complex Analysis
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Lecture 23-24 Complex Functions
Dr Trilok Mathur, Assistant Professor, Department of Mathematics
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• Function of a complex variable Let S be a set complex numbers. A function f defined
on S is a rule that assigns to each z in S a complex number w.
3
S
Complex numbers
f
S’
Complex numbers
z
The domain of definition of f The range of f
w
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Suppose that w = u + iv is the value of a function f at z = x + iy, so that
Thus each of real number u and v depends on the real variables x and y, meaning that
Similarly if the polar coordinates r and θ, instead of x and y, are used, we get
4
( )u iv f x iy
( ) ( , ) ( , )f z u x y iv x y
( ) ( , ) ( , )f z u r iv r
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• Properties of a real-valued function of a real variable are often exhibited by the graph of the function. But when w = f(z), where z and w are complex, no such convenient graphical representation is available because each of the numbers z and w is located in a plane rather than a line.
• We can display some information about the function by indicating pairs of corresponding points z = (x, y) and w = (u, v). To do this, it is usually easiest to draw the z and w planes separately.
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x u
y v
z-plane w-plane
domain of definition
range
w = f(z)
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• Example: If f (z) = z2, then
case #1: case #2:
7
2 2 2( ) ( ) 2f z x iy x y i xy
2 2( , ) ; ( , ) 2u x y x y v x y xy
z x iy
iz re 2 2 2 2 2( ) ( ) cos2 sin 2i if z re r e r ir
2 2( , ) cos2 ; ( , ) sin 2u r r v r r
When v = 0, f is a real-valued function.
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• Example: A real-valued function is used to illustrate some
important concepts later in this chapter is • Polynomial function:
where n is zero or a positive integer and a0, a1, …an
are complex constants, an is not 0;
• Rational function: the quotients P(z)/Q(z) of polynomials
8
2 2 2( ) | | 0f z z x y i
20 1 2( ) ... n
nP z a a z a z a z
The domain of definition is the entire z-plane
The domain of definition is Q(z)≠0
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9
Suppose the function f is defined in some neighborhood of z0, except possibly at z0 itself. Then f is said to possess a limit at z0, written
0)(lim0
wzfzz
δz-z-wzf
δ,
00 0)(
00
whenever
that such a given for if
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10
meaning the point w = f (z) can be made arbitrarily chose to w0 if we choose the point z close enough to z0 but distinct from it.
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,,
),,(),()(
000000 ivuwiyxz
yxivyxuzf
Let Thm 1
0)(),(
0)(),(
0
)(lim)(
)(lim)(
)(lim
0,0
0,0
0
vx,yvii
ux,yui
wzf
yxyx
yxyx
zz
Then
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Then .
, Let
0
0
0
0
lim
)(lim
W F(z)
wzf
zz
zz
Thm 2
.)]()(lim 000
WwzFzfzz
[ (i)
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.W)]()([ lim)( 000
wzFzfiizz
.0 Wif ,W)(
)( lim)( 0
0
0
0
w
zF
zfiii
zz
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• Example Show that in the open disk | z | < 1, then
Proof:
( ) / 2f z iz
1lim ( )
2z
if z
| || 1| | 1|| ( ) | | |
2 2 2 2 2
i iz i i z zf z
0, 2 , . .s t
when 0 | 1| ( 2 )z
| 1|0 | ( ) |
2 2
z if z
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15
• Example If then the limit does not exist.
( ,0)z x
( )z
f zz
0
lim ( )z
f z
0
0lim 1
0x
x i
x i
(0, )z y0
0lim 1
0y
iy
iy
≠
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The point at infinity is denoted by
, and the complex plane together
with the point at infinity is called
the Extended complex Plane.
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Riemann Sphere & Stereographic Projection
N: The North pole
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18
• The -Neighborhood of Infinity
O x
y
R1
R2
When the radius R is large enough
i.e. for each small positive number
The region of |z| > R = is called the -Neighborhood of Infinity(∞)
0
1R
1
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.0)(
1lim)(lim.1
00
zf
zfzzzz
00
0
1lim)(lim.2 w
zfwzf
zz
01
1lim)(lim.3
0
zf
zfzz
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δz-zzf
ei
z-z zf
a
zfzz
0
0
00)(
1..
01
)(
0
,0)(lim)10
whenever
whenever
that such
eachfor
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.0)(
1lim)(lim
00
zf
zfzzzz
Thus,
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22
A function f is continuous at a point z0 if
meaning that 1. the function f has a limit at point z0 and 2. the limit is equal to the value of f (z0)
00lim ( ) ( )
z zf z f z
For a given positive number ε, there exists a positive number δ, s.t.
0| |z z When ⇒ 0| ( ) ( ) |f z f z
00 | | ?z z
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23
The function f(z) is said to be
continuous in a region R if it is
continuous at all points of the
region R.
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24
Theorem 1. A composition of two continuous functions is itself
continuous. Theorem 2. If f (z) = u(x, y) + iv(x, y), then f (z) is continuous iff Re(f (z)) = u(x, y) and Im(f (z)) = v(x, y)
are continuous functions.
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25
.continuousallare
(c)
(b)
(a)
then ,continuous are and If 3.Theorem
0)(,)(
)(
)()(
)()(
)()(
zgzg
zf
zgzf
zgzf
zgzf
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26
Theorem 4. If a function f (z) is continuous and nonzero at a point z0, then f (z) ≠ 0 throughout some neighborhood of that point.
Proof 0
0lim ( ) ( ) 0z z
f z f z
0| ( ) |0, 0, . .
2
f zs t
0| |z z When 0
0
| ( ) || ( ) ( ) |
2
f zf z f z
f(z0) f(z)
If f (z) = 0, then 00
| ( ) || ( ) |
2
f zf z
Contradiction!
0| ( ) |f z
Why?
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27
Theorem 5. If a function f is continuous throughout a region R that is both closed and bounded, there exists a nonnegative real number M such that
where equality holds for at least one such z.
| ( ) |f z M for all points z in R
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Example: Discuss the continuity of f (z) at z = 0 if
zz zf ii
z
zzf i
Re
Re
1)()(
1)()(
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Derivatives: Let f (z) be a function defined
on a set S and S contains a nbd of z0. Then
derivative of f (z) at z0, written as
is defined by the equation:
),( 0zf
exists. RHS on limit the provided
)1(,)()(
lim)(0
00
0 zz
zfzfzf
zz
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exists. at derivative its if
at abledifferenti be to said is function The
0
0)(
z
zzf
z
zfzzfzf
zz-z
z
)()(lim)(
,
00
00
0 to reduces (1) then If
30
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ablityDifferenti Continuity
Continuityablity Differenti :Problem
0
00
0
)()(lim)(
)(
0 zz
zfzfzf
zzf
zz
at abledifferenti is Let :Proof
)()(lim 00
zfzfzz
Now
)(
)()(lim 0
0
0
0
zzzz
zfzfzz
31
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00)('
)(lim)()(
lim
0
00
0
00
zf
zzzz
zfzfzzzz
)()(lim 00
zfzfzz
0)( zzf at continuousis
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Continuity Differentiability
To show this consider the function
f (z) =z2 = x2 + y2 = u(x, y)+ i v(x, y)
everywhere continuous is hence
,everywhere continuous are and Since
)(
.0),(,),( 22
zf
vu
yxvyxyxu
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0
00
0
2
0
2
0
0
0
)()(
,
zz
zzzz
zz
zz
zz
zf zf
zz
have we For
0
0000 zz
zzzzzzzz
0
000 )()(
zz
zzzzzz
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zzzz
zzz
00 ,.
yix
yixzz
.0
x
y
C1
C2
20
10
,
,
Czz
Czz
paththealong
paththealong
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unique.notis
then if Thus,
0
0
0
)()(lim
)00(
0 zz
zfzf
,,z
zz
else. whereno
and origin the at abledifferenti is
then When
)(
.0)()(
lim
)00(
00
0
0
0
zf
zzz
zfzf
,,z
zz
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)()( zfdz
d zf 1.
,0.2 cdz
d
,1.3 zdz
d
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,.4 1 nn nzzdz
d
zfdz
dczcf
dz
d .5
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)()()()(.6 zgzfzgzfdz
d
)()()()()()(.7 zgzfzgzfzgzfdz
d
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0)(
,))((
)()()()(
)(
)(.8
2
zg
zg
zfzgzfzg
zg
zf
dz
d
if
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Let F(z) = g(f (z)), and assume that f (z) is
differentiable at z0 & g is differentiable at
f (z0), then F(z) is differentiable at z0 and
)())(()( 000 zfzfgzF
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dz
dw
dw
dW
dz
dWzFW
w gW zfw
rule Chainby hence
and Let Ex.
),(
)()(
. pointany at
exist not does thatshow (a)8Q
z
zfzzf )(,)(:
0
0
0
0
0
0
0
)()(
,
zz
zz
zz
zz
zz
zfzf
zz
then Let :Solution
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yix
yix
z
z
z
zfzzf
)()( 00
any whereexist not does)(
1
1
)()(
0
lim
2
1
00
zf
Calong
Calong
z
zfzzf
z
x
y
C1
C2
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exist. NOT does thatShow
by defined function a be Let Q.9
)0(
0,0
0,
)(
2
f
z
zz
z
zf
f
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000
)(
)()()(
iyxz
zf
yx,ivyx,uzf
point a at exists that and
that Suppose
equations-CR thesatisfy they and
at exist must and
sderivative partial order-first the Then
)( 00 y,xv,v,uu yxyx
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).,(),()('
)(
00000
0
yxivyxuzf
zf
xx
as writtenbe can
Also,
Proof
)1(......)()(
lim)(
)(
00
00
0
z
zfzzfzf
zzf
z
at abledifferenti is Since
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),(),()(
,
00000
000
yxviyxuzf
yixz
iyxziyxz
that Note
),(
),()(
00
000
yyxxiv
yyxxuzzf
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yix
yxvyyxxvi
yix
yxuyyxxu
yx
zf
),(),(
),(),(
)0,0(),(
lim
)(
0000
0000
0
givesEq. )1(
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x
y
C1
C2
2
0000
1
0000
0
),,(),(
),()(
)(
C
yxvyxui
C
yxviyxu
zf
yy
xx
along
along
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???WHY
),()(
),,(,
000
00
yxatviuzfand
yxatvuvu
xx
xyyx
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:thatsuch
point the of oodneighbourh
some in throughout defined function
any be Let
000
)()()(
iyxz
x, yi v x,y u zf
,,,,)( 0zvvuui yxyx of nbd. that in exist
),(,,,)( 00 yxvvuuii yxyx at continuous are
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).,(,
)(
00 yxvuvu
iii
xyyx atequations,-CR the
satisfy sderivative partial order first the
.)( 0zzf at exists Then
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satisfy
they and( at exist sderivative
partial the Then point givenany at
abledifferenti be Let
),,,
.
)()(
00
000
θ,rvvuu
erz
θr,vir,θ uf(z)
rr
iθ
ur
vvr
u rr
1,
1
oorrri viuezf
,0 )(
and
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find out the points where the function is differentiable. Also find
,z zf 2)(
function the For1: Example
)(zf
ivuxyiyx z zf 2)( 222
Consider:Solution
xyyxvyxyxu 2),(,),( 22
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xvyv
yuxu
yx
yx
2,2
,2,2
xyyx vuvu &
x, y all for satisfied are equations- CR (i)
x, y
vvuu yxyx
all
for continuous are and (ii) ,,,
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and
pointany at abledifferenti is z,zzf 2)(
zyixivuzf xx 222)(
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find out the points where the function is differentiable. Also find
,)(2
zzf
function the For:2 Example
)(zf
222)( yxz zf Consider
0),(&),( 22 yxvyxyxu
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,0,0,2,2 yxyx vvyuxu
y. x 0
have must wethen
satisfied, are equations- CR If
0)0,0()0,0()0(
)00()(
xx ivu f
,zf
Further else. whereno and
atonly abledifferenti is
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by defined function
the of partsimaginary & real
the denote Let :Q.672 Page
f
u & v
0,0
0,)(
2
z
zz
zzf
. at abledifferenti NOT is although
at satisfied are equations-CR thatShow
)00(
)00(
,f
,
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z
z
z
zzz
z
z-fzzf zf
zf.z zf
zz
z
00
0
limlim
)()(lim)(
)()(
have We
IMethod
exist?Does Let Q.
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point.any at exist not does
paththealong
paththealong
)(
1
1
lim
2
1
)0,0(,
zf
C
C
yix
yixyx
z = x + i y
c1
c1
c2
c2
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xyyx
yx
yx
vuvu
vv
uu
-yx, vu
x-iyz zf
,
1,0
,0,1
)(
Thus
II-Method
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point.any at abledifferenti not is
satisfied not areequationsC.R.
)(zf
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).()(
)(.)(
zfzf
zfezf z
and
find and abledifferenti is
thatShow Let 2b.Q
yevyeu
yiye
eezf
xx
x
iyxz
sin,cos
)sin(cos
)(
:Solution
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.cos
,sin
,sin
,cos
yev
yev
yeu
yeu
xy
xx
xy
xx
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. point
any at continuous are
(1)
Clearly
)(
,,,)2(
&
x,y
vvuu
vuvu
yxyx
xyyx
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ziyx
xx
xx
eee
yeiye
ivuzf
zf
.
sincos
)(
)(
and exists
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yeV
y-eU
ViU
yeiy-e
zf zF
zf
x
-x
x-x
sin
,cos
sincos
)()(
)(
(say)
Let
: find To
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.cos
,sin
,sin
,cos
yeV
yeV
yeU
yeU
xy
xx
xy
xx
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point
any at continuous
are
Thus,
)(
,,,)2(
&)1(
yx,
VVUU
VUVU
yxyx
xyyx
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z
iyx
xx
xx
e
ee
yeiye
iVUzf zF
zF
& exists
.
)sin(cos
)()(
)(
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if point a at
analytic be to said is function
0
)(
z
zfA
and at abledifferenti is ,)()( 0zzfi
.
)()(
0z
zfii
of oodneighbourh
some in abledifferenti is
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. set the of point each
at abledifferenti is f if set open
an inanalytic is functionA
S
S
zf )(
y.analyticitimply not doesablity Differenti
:Remark
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:Ex2
)( z zf
f (z) is differentiable at origin and no where else.
But f (z) is not analytic at the origin as it is not differentiable in any neighborhood of origin.
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. in throughout constant is then
, domain a in everywhere If
:Theorem
Dzf
Dzf
)(
0)(
Entire function: A function f(z) is said to
be an Entire function if f(z) is analytic
at each point in the entire finite plane.
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Singular Point: Let a function f (z) is, not analytic at a point z0, but analytic at some point in every neighbourhood of z0.
Then z0 is called a singularity of f(z).
Example: Every polynomial is an entire
function.
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z
zf 1
)()1(
Examples
. ofy singularit a is )(0 zf z
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2)()2( z zf
anywhereanalytic not is zf )(
point singular no has )(zf
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