MATH V3007 Complex Variables - Solution to HW4

Problem 1 (2.11 in Stein-Shakarchi)

a) From the hint:

0 = − 1

2πi

∫ 2π

0

f(Reiθ)

Reiθ − R2

z̄

iReiθdθ

From Cauchy’s integral formula:

f(z) =1

2πi

∫ 2π

0

f(Reiθ)

Reiθ − ziReiθdθ

Adding the two gives:

f(z) =1

2π

∫ 2π

0

f(Reiθ)

(Reiθ

Reiθ − z+

z̄

Re−iθ − z̄

)dθ

f(z) =1

2π

∫ 2π

0

f(Reiθ)1

2

(Reiθ + z

Reiθ − z+Re−iθ + z̄

Re−iθ − z̄

)dθ

f(z) =1

2π

∫ 2π

0

f(Reiθ)Re

(Reiθ + z

Reiθ − z

)dθ

b)Multiplying and dividing by the conjugate of the denominator we get:

Reit + r

Reit − r=R2 − 2irR sin t− r2

R2 − 2rR cos t+ r2

Re

(Reit + r

Reit − r

)=

R2 − r2

R2 − 2rR cos t+ r2

Problem 2 (2.14 in Stein-Shakarchi)

We can write the Taylor expansion of f around 0 as f(z) =∑∞

n=0 anzn and its Laurent

expansion around z0 as∑∞

n=−k bn(z − z0)n; they both converge in D. We can relate thesetwo series:

an = bn +−1∑j=−k

(−1)jdj(n− j − 1)(n− j − 2)...(n+ 1)

(−j − 1)!(z0)n−j+1

1

When we take the limit n → ∞ of an/an+1, the highest order contribution comes from theterms:

limn→∞

d−k(n+k−1)(n+k−2)...(n+1)

(k−1)!zn+k+10

d−k(n+k)(n+k)...(n+2)

(k−1)!zn+k+10

= z0

Problem 3 (2.15 in Stein-Shakarchi)

Define the function F as suggested in the hint. We want to show that F is entire andbounded, which would imply that it is constant. If |z| = 1 then |f(z) = 1 so:

F (z) =1

f(1/z̄)=

1

f(z)= f(z)

So F is everywhere continuous. Since f(1/z̄) is holomorphic and nonzero in D, F = 1

f(1/z̄)

will be holomorphic in C − D. To show that F is also holomorphic for |z| = 1, we can useMorera’s theorem. Take any triangle whose interior intersects the unit circle nontrivially,and break it into smaller triangles. Some triangles will be in D or C − D; the integral of Fover these is 0 since F is holomorphic in these domains. Some triangles will still intersectthe unit circle, but they can be made arbitrarily small, so the integral of F over them is 0as well. This completes the proof that F is entire. It is bounded because f is bounded andnonzero in D.

Problem 4 (3.2 in Stein-Shakarchi)

Integrate over the upper semicircular contour; the integral over the semicircular part is 0since the degree of the denominator is greater than 2. Therefore the desired integral is justthe sum of all residues that lie in the upper semicircular contour. The poles are the 4-throots of -1: eiπ/4, ei3π/4, ei5π/4, ei7π/4. Only the first two are inside the contour, so:∫ ∞

−∞

dx

1 + x4= 2πi

(1

4e−iπ/4 +

1

4e−i3π/4

)=

π√2

Problem 5 (3.3 in Stein-Shakarchi)

The desired integral is the real part of: ∫ ∞−∞

eix

x2 + a2

We integrate over the upper semicircular contour. The integral over the semicircular vanishesby Jordan’s lemma, so the contour integral is equal to the residue of the pole ia:∫ ∞

−∞

eix

x2 + a2= 2πiResai(f) =

πe−a

a

This is already real, so its real part is itself.

2

Problem 6 (3.5 in Stein-Shakarchi)

We choose the integration contour based on the sign of ζ. If ζ < 0, we are forced to choosethe upper semicircle. If ζ > 0, we are forced to choose the lower semicircle. If ζ = 0, eithersemicircle works. We’ll treat the case ζ < 0, the other two are analogous. Again, usingJordan’s lemma the integral over the semicircular part vanishes, so the desired integral isequal to the residue at the pole i:∫ ∞

−∞

e−2πxζ

(1 + x2)2dx = 2πie2πζ

[πiζ

2+

1

4i

]=π

2e−2π|ζ| (1 + 2π|ζ|)

3