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Material Type: Exam; Class: Complex Analysis; Subject: Mathematics; University: University of Massachusetts - Amherst; Term: Spring 2006;
Typology: Exams
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= (^) 1+^1 i
n=0(−1)n^
(z− 1 1+i
∑∞ n=
(−1)n (1+i)n+1^ (z^ −^ 1)
n. The radius of convergence is |1 + i| = √2.
(a)
C
e^2 iz^ + z^2 + 3 (z − 2)^3
dz, where C = {z : |z| = 3} transversed counterclockwise.
Answer: ∫ Set f (z) = e^2 iz^ + z^2 + 3. Then
C
e^2 iz^ + z^2 + 3 (z − 2)^3
dz = 2πi ·
f (2)(2) 2!
= πi((−4)e^4 i^ + 2) = π[sin(4) + i(2 − 4 cos(4))].
(b) I :=
C
dz z(1 − z)
, along the path C consisting of the line segments C 1 , C 2 , C 3 , where C 1 goes from −2 to −i, C 2 goes from −i to 1 + i, and C 3 goes from 1 + i to 2.
Answer: Write (^) z(1^1 −z) = (^1) z − (^) z−^11. We find primitives, for each summand, defined in an open set containing C, by setting
f (z) = log(z) with arg(z) ∈ (− 3 π/ 2 , π/2), g(z) = log(z − 1) with arg(z − 1) ∈ (−π/ 2 , 3 π/2).
Thus, I =
C f^
′(z) − g′(z)dz = [f − g]^2 − 2 =^ {ln(2)^ −^ ln(1)} − {(ln(2)^ −^ iπ)^ − (ln(3) + iπ)} = ln(3) + 2πi.
(a) Let f (z) and g(z) be two entire functions, such that |f (z)| ≤ |g(z)|, for every complex number z. Then f (z) = cg(z), for some constant c. Answer: True. If f is identically 0, take c = 0. Otherwise, g(z) is not identically zero. The function h(z) := f g^ ((zz)) has isolated singularities, since the zeros of g(z) are isolated. Furthermore, h(z) is bounded, by assumption. Hence, all the singularities of h(z) are removeable and h(z) extends to a bounded entire function, which must be constant, by Liouville’s Theorem. (b) Let f and g be holomorphic functions on the upper-half-plane H := {x + iy : y > 0 }. If f ( (^) ni ) = g( (^) ni ), for all integers n ≥ 1, then f = g. Answer: False. Take f (z) = 1 and g(z) = e^2 π/z^. (c) Let U ⊂ C be an open connected subset. If both f (z) and f (z) are holomor- phic functions on U, then f is constant in U. Answer: True. Everybody did this part correctly.
Answer: f must map the boundary to the boundary preserving the orientation. Being a conformal map (preserving angles), f must also map the set of vertices { 1 , − 1 } to itself in a one-to-one fashion. Hence, f (−1) = 1. The orientation determines the cyclic order of the boundary points {− 1 , 0 , 1 }, which must hence be preserved. Consequently, λ := f (0) must belong to the upper-half of the unit circle. We can proceed to argue in two ways. We can prove that an automorphism f of U exists, for λ = eiθ^ , and for any choice of θ ∈ (0, π), and then compute it. Or we can compute f , for every such choice, and then prove that it indeed maps U onto itself. First method: Let W := {x + iy : x > 0 , y > 0 } be the first quadrant. The linear fractional transformation h(z) = −
(z+ z− 1
, sending (− 1 , 0 , 1) to (0, 1 , ∞), takes U onto W. The linear fractional transformation τ (z) = (^) zi maps W onto itself and interchanges the x and y axis. Now W is invariant under multiplication by any positive real number t. So the composition f (z) = (h−^1 ◦tτ ◦h)(z) = (− (1^1 −−titi))zz+(1++(−1+ti)ti) is the l.f.t we were looking for, with λ = f (0) = − 1+1+titi. Second method: The l.f.t taking (− 1 , 0 , 1) to (1, λ, −1) is given by f (z) = (^) λzz−−λ 1. It maps the real line to the unit circle, since there is a unique circle or line through three distinct points of the plane and { 1 , λ, − 1 } are on the unit circle. Observe that f 2 = id. Hence, f takes the unit circle to the real line. It follows that f maps U either onto itself, or onto CP^1 \ U. We constructed f to preserve the orientation of the boundary (by making sure it preserves the cyclic order). Hence, f maps U onto itself. Alternatively, check that f −^1 (∞) = (^1) λ does not belong to U. So f (U) = U.
∣∣ f (n)(a)
n! · MR (R − |a|)n+^
for any complex number a, such that |a| < R. Answer: First Method: (Estimate the integral in Cauchy’s Formula) We have
f (n)(a) =
n! 2 πi
|z|=R
f (z) (z − a)n+^
dz.
Thus,
f (n)(a)
≤ 2 nπ!
∫ (^2) π 0
|f (z)| |z−a|n+
iReiθ
dθ ≤ n 2 !πR
∫ (^2) π 0
M (R−|a|)n+1^ dθ^ =^
n!M R (R−|a|)n+ Second method: Use Cauchy’s estimate with the circle Ca of radius R−|a| centered at a. f is holomorphic on that disk, since that disk is contained in the disk of radius R centered at 0, We need however a bound for |f | on the circle Ca. By the Maximum principle, we can take M as an upper bound (we did not cover the Maximum principle before the exam, so some students used this method without explaining this point). We get ∣∣ f (n)(a)
n!M (R − |a|)n^
n!MR (R − |a|)n+^