solution manual for saff and snider complex analysis
MATH V3007 Complex Variables - Solution to HW4
Problem 1 (2.11 in Stein-Shakarchi) a) From the hint: 1 0=− 2πi
2π
Z 0
f (Reiθ ) iReiθ dθ R2 iθ Re − z¯
From Cauchy’s integral formula: 1 f (z) = 2πi
Z 0
2π
f (Reiθ ) iReiθ dθ Reiθ − z
Adding the two gives: Z 2π 1 Reiθ z¯ iθ f (z) = f (Re ) + dθ 2π 0 Reiθ − z Re−iθ − z¯ iθ Z 2π Re + z Re−iθ + z¯ 1 iθ 1 f (Re ) + f (z) = dθ 2π 0 2 Reiθ − z Re−iθ − z¯ iθ Z 2π Re + z 1 iθ f (Re )Re dθ f (z) = 2π 0 Reiθ − z b)Multiplying and dividing by the conjugate of the denominator we get: Reit + r R2 − 2irR sin t − r2 = Reit − r R2 − 2rR cos t + r2 it Re + r R2 − r 2 = Re Reit − r R2 − 2rR cos t + r2
Problem 2 (2.14 in Stein-Shakarchi) P∞ n We can write the TaylorPexpansion of f around 0 as f (z) = n=0 an z and its Laurent n expansion around z0 as ∞ n=−k bn (z − z0 ) ; they both converge in D. We can relate these two series: −1 X (n − j − 1)(n − j − 2)...(n + 1) an = b n + (−1)j dj (−j − 1)!(z0 )n−j+1 j=−k 1
When we take the limit n → ∞ of an /an+1 , the highest order contribution comes from the terms: d−k (n+k−1)(n+k−2)...(n+1) (k−1)!z0n+k+1 lim = z0 n→∞ d−k (n+k)(n+k)...(n+2) n+k+1 (k−1)!z 0
Problem 3 (2.15 in Stein-Shakarchi) Define the function F as suggested in the hint. We want to show that F is entire and bounded, which would imply that it is constant. If |z| = 1 then |f (z) = 1 so: F (z) =
1 f (1/¯ z)
=
1 f (z)
= f (z)
So F is everywhere continuous. Since f (1/¯ z ) is holomorphic and nonzero in D, F =
1 f (1/¯ z)
will be holomorphic in C − D. To show that F is also holomorphic for |z| = 1, we can use Morera’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 F over these is 0 since F is holomorphic in these domains. Some triangles will still intersect the unit circle, but they can be made arbitrarily small, so the integral of F over them is 0 as well. This completes the proof that F is entire. It is bounded because f is bounded and nonzero in D.
Problem 4 (3.2 in Stein-Shakarchi) Integrate over the upper semicircular contour; the integral over the semicircular part is 0 since the degree of the denominator is greater than 2. Therefore the desired integral is just the sum of all residues that lie in the upper semicircular contour. The poles are the 4-th roots of -1: eiπ/4 , ei3π/4 , ei5π/4 , ei7π/4 . Only the first two are inside the contour, so: Z ∞ 1 −iπ/4 1 −i3π/4 π dx √ = 2πi e + e = 4 4 4 2 −∞ 1 + x
Problem 5 (3.3 in Stein-Shakarchi) The desired integral is the real part of: Z
∞
−∞
eix x 2 + a2
We integrate over the upper semicircular contour. The integral over the semicircular vanishes by Jordan’s lemma, so the contour integral is equal to the residue of the pole ia: Z ∞ eix πe−a = 2πiRes (f ) = ai 2 2 a −∞ x + 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 choose the upper semicircle. If ζ > 0, we are forced to choose the lower semicircle. If ζ = 0, either semicircle works. We’ll treat the case ζ < 0, the other two are analogous. Again, using Jordan’s lemma the integral over the semicircular part vanishes, so the desired integral is equal to the residue at the pole i: Z ∞ −2πxζ 1 π e 2πζ πiζ + = e−2π|ζ| (1 + 2π|ζ|) dx = 2πie 2 2 2 4i 2 −∞ (1 + x )
