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HOMEWORK 4.
Due Monday, March 23.
- Suppose that u, v are real-valued functions having continuous partial deriva-
tives of first order in a neighborhood of z 0 = x 0 + iy 0. Prove that f = u + iv is
differentiable (with respect to z) at z 0 iff
rlim→ 0 1 πr^2
C(z 0 ,r)
f (z) dz = 0.
Solution: By Taylor’s formula
u(z 0 + h) = u(z 0 ) + ux(z 0 )Reh + uy (z 0 )Imh + h o(1), and similarly for v. Obtain
f (z 0 +h) = f (z 0 )+fx(z 0 )Reh+fy (z 0 )Imh+h o(1). If h = reit^ and γ(t) = z 0 +reit then ∫
γ
f (z)dz =
∫ (^2) π
0
f (z 0 + reit)ireit^ dt
= fx(z 0 )
∫ (^2) π
0
r cos t ireit^ dt + fy (z 0 )
∫ (^2) π
0
r sin t ireit^ dt + o(r)r
= [fx(z 0 ) + ify (z 0 )]πr^2 + o(r^2 ).
Consequently, the integral on the left converges to 0 iff fx(z 0 ) + ify (z 0 ) = 0
which is equivalent to C-R equations.
- Suppose that F is analytic in a convex domain G and ReF ′(z) > 0 for any z ∈ G. Prove or disprove: F is univalent in G (i.e., F (z 1 ) 6 = F (z 2 ) for any
z 1 6 = z 2 ).
Solution: Suppose z 1 6 = z 2 and F (z 1 ) = F (z 2 ). Define β(t) = z 1 + t(z 2 −
z 1 ), γ(t) = F ′[β(t)], 0 ≤ t ≤ 1. Since ReF ′(z) > 0, 0 ∈ {/ γ}. OTOH, 0 =
β F^
′(z) dz = ∫^1 0 F^
′[z 1 + t(z 2 − z 1 )] (z 2 − z 1 ) dt so ∫^1 0 F^
′[z 1 + t(z 2 − z 1 )] dt = 0.
Further,
0 F^
′[z 1 + t(z 2 − z 1 )] dt = lim ∑ k F^
′[z 1 + tk(z 2 − z 1 )] ∆tk and the last
sum represents a convex combination of points on {γ} so 0 belongs to the closed
convex hull of {γ}.
- Do Problem 4 on page 74 of the text.
Solution: Use (36.1) in Bartle’s text with xj = rj^ , yj = aj , n = 0, sj = ∑j
∑k=0^ ak,^ s−^1 = 0, to obtain m j=0 aj^ rj^ =^ smrm+1^ +^
∑m j=0 sj^ (rj^ −^ rj+1). Also 0 = A[
∑m
∑ j=0(rj+1^ −^ rj^ ) + 1^ −^ rm+1]. Adding, m j=0 aj^ rj^ =^
∑m
∑ j=0(sj^ −^ A)(rj^ −^ rj+1) + (sm^ −^ A)rm+1^ +^ A.^ When^ m^ → ∞, ∞ j=0 aj^ r
j (^) = ∑∞ j=0(sj^ −^ A)r
j (^) (1 − r) + A. Now let r ↑ 1.
- Do Problem 10 on page 75 of the text.
Solution: z
z(z^2 + 4)
z
z + 2i
z − 2 i
. When 0 < r < 2, the
integral equals 1 8
(2 · 2 πi + 0 + 0) = πi 2
. When r > 2, the integral equals 1 8 (2^ ·^2 πi^ + 3^ ·^2 πi^ + 3^ ·^2 πi) = 2πi.
- Compute the sums of the following series: (a)
n=
cos nϕ n , 0^ < ϕ^ ≤^ π;
(b)
n=
sin(2n + 1)ϕ 2 n + 1 , 0^ < ϕ < π.
Solution: (a) For |z| < 1, − log(1 − z) =
n≥ 1 zn/n. Write^ z^ =^ reit^ and take the real parts to get − ln | 1 − reit| =
n≥ 1 r
n (^) cos nt/n. By Problem 3, we can
take the limits r ↑ 1. Result: − ln |2 sin(t/2)|. (b) Use the imaginary parts of 1 2
ln 1 +^ z 1 − z
n≥ 0
z^2 n+ 2 n + 1
. Result: π/4.
- Let
n=0 cnzn^ be the power series representation for the function 1/(1^ −^ z^ − z^2 ). Find the coefficients cn and the radius of convergence of the series.
Solution: Write f (z) =
n≥ 0 anzn,^ an^ =^ |an|eitn^ , and^ γ(t) =^ ρ(t)eiϕ(t), a ≤ t ≤ b. Then ∫
γ
f f ′^ dz =
∫ (^) b
a
f (γ(t))f ′(γ(t))γ′(t) dt
∫ (^) b
a
n≥ 0
m≥ 0
amnanρm+n−^1 e−imϕei(n−1)ϕ[ρ′eiϕ^ + ρeiϕiϕ′]
∫ (^) b
a
n≥ 0
m≥ 0
|am||an|nei(tn−tm)+(n−m)ϕρm+n−^1 [ρ′^ + iρϕ′].
Writing αmn = tn − tm + (n − m)ϕ, the real part equals ∫ (^) b
a
n≥ 0
m≥ 0
|am||an|nρm+n−^1 [ρ′^ cos(αmn) − ϕ′^ sin(αmn)].
Notice
n≥ 0
m≥ 0 cmn^ =^
m≥ 0 cmm^ +
m≥ 0
∑^ n<m(cmn^ +cnm). Here m≥ 0 cmm^ =^
∫ (^) b a
m≥ 0 |am|^2 mρ^2 m−^1 [ρ′] = (1/2)^
∫ (^) b a [
m≥ 0 |am|^2 mρ^2 m]′^ = 0. Remains: ∫ (^) b
a
m≥ 0
n<m
|am||an|ρm+n−^1 [(m + n)ρ′^ cos(αmn) − (n − m)ρϕ′^ sin(αmn)] =
∫ (^) b
a
m≥ 0
n<m
[ρm+n|am||an| cos(αmn)]′^ = 0.
If f is not analytic in a disk, cover {γ} with a finite number of disks so
that f is analytic in each. Now {γ} = {γ 1 } ∪ · · · ∪ {γn} and the integrand has
real part with a primitive F , so Re(
{γk } ...) =^ F^ (bk)^ −^ F^ (ak) and the result follows by adding the pieces.
