Solutions to Midterm Exam - Complex Analysis | MATH 621, Exams of Mathematics

Material Type: Exam; Class: Complex Analysis; Subject: Mathematics; University: University of Massachusetts - Amherst; Term: Spring 2006;

Typology: Exams

Pre 2010

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Math 621 Solution of the Midterm Spring 2006
1. (10 points) Find the power series expansion of f(z) = 1
z+i, centered at z= 1.
What is its radius of convergence?
Answer: 1
z+i=1
(z1)+1+i=1
1+i
1
(z1
1+i)+1 =1
1+iP
n=0(1)nz1
1+in=
P
n=0
(1)n
(1+i)n+1 (z1)n.The radius of convergence is |1 + i|=2.
2. (18 points) Compute the following integrals. Justify all your steps.
(a) ZC
e2iz +z2+ 3
(z2)3dz, where C={z:|z|= 3}transversed counterclockwise.
Answer: Set f(z) = e2iz +z2+ 3. Then
ZC
e2iz +z2+ 3
(z2)3dz = 2πi ·f(2) (2)
2! =πi((4)e4i+ 2) = π[sin(4) + i(2 4 cos(4))].
(b) I:= ZC
dz
z(1 z),along the path Cconsisting of the line segments C1,C2,
C3, where C1goes from 2 to i,C2goes from ito 1 + i, and C3goes from
1 + ito 2.
Answer: Write 1
z(1z)=1
z1
z1. We find primitives, for each summand, defined
in an open set containing C, by setting
f(z) = log(z) with arg(z)(3π/2, π /2),
g(z) = log(z1) with arg(z1) (π/2,3π/2).
Thus, I=RCf0(z)g0(z)dz = [fg]2
2={ln(2) ln(1)} {(ln(2) )
(ln(3) + )}= ln(3) + 2πi.
3. (18 points) Determine if the following statements are true or false. If true, prove
the statement. If false, give a counter example.
(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 fis identically 0, take c= 0. Otherwise, g(z) is not
identically zero. The function h(z) := f(z)
g(z)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 fand gbe holomorphic functions on the upper-half-plane
H:= {x+iy :y > 0}. If f(i
n) = g(i
n), for all integers n1, then f=g.
Answer: False. Take f(z) = 1 and g(z) = e2π/z .
(c) Let UCbe an open connected subset. If both f(z) and f(z) are holomor-
phic functions on U, then fis constant in U.
Answer: True. Everybody did this part correctly.
4. (18 points) Find all linear fractional transformations, which map the upper half of
the unit disk U:= {z:|z|<1 and Im(z)>0}onto itself, and satisfy f(1) = 1.
Justify your answer. Hint: There are infinitely many such maps.
1
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Math 621 Solution of the Midterm Spring 2006

  1. (10 points) Find the power series expansion of f (z) = (^) z+^1 i , centered at z = 1. What is its radius of convergence? Answer: (^) z^1 +i = (^) (z−1)+1+^1 i = (^) 1+^1 i (z−^11 1+i )+^

= (^) 1+^1 i

n=0(−1)n^

(z− 1 1+i

)n

∑∞ n=

(−1)n (1+i)n+1^ (z^ −^ 1)

n. The radius of convergence is |1 + i| = √2.

  1. (18 points) Compute the following integrals. Justify all your steps.

(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.

  1. (18 points) Determine if the following statements are true or false. If true, prove the statement. If false, give a counter example.

(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.

  1. (18 points) Find all linear fractional transformations, which map the upper half of the unit disk U := {z : |z| < 1 and Im(z) > 0 } onto itself, and satisfy f (1) = −1. Justify your answer. Hint: There are infinitely many such maps.

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.

  1. (18 points) Let f be a holomorphic function on the closed disk {z : |z| ≤ R}. Set M := max{|f (z)| : |z| = R}. a) Prove the inequality

∣∣ 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+^