MIDTERM COMPLEX ANALYSIS
No credit will be given for an answer without reasoning.
If f is analytic inside and on the simple closed positively oriented contourΓ and if z0 is insideΓ. The (generalized) Cauchy integral formula:
f(n)(z0) = n!
2πi Z
Γ
f(z) (z − z0)n+1dz.
1.
(1) [5%] Computearg i.
(2) [5%] Let x, y be real numbers. Show that|ex+iy| = ex. 2.
(1) [5%] Calculate(1 −√ 3i)3(√
3 + i)2.
(2) [5%] Represent the number (1−i)1 2 in polar form.
3. Let f(z) = z4− 4z3+ 6z2− 4z + 5.
(1) [5%] Check that i is a root of f(z) = 0.
(2) [5%] Find all other roots.
4.
(1) [5%] Find all values of z such that ez = −4.
(2) [5%] Find the principal value of(1 + i)πi. 5.
(1) [5%] Let D be the rectangle T in z plane with vertices 1, 1 + πi, −1 + πi, −1. Describe the image of T in w plane under the map w= ez.
(2) [5%] Show that if v is a harmonic conjugate of u in a domain D, then uv is harmonic in D.
6.
(1) [5%] Give a parametrization of the following contour starting from −i to i with parameter 0 ≤ t ≤ 1.
-4 -3 -2 -1 0 1 2 3 4
-2 -1 0 1 2
1
2 MIDTERM COMPLEX ANALYSIS
(2) [5%] Compute the line integral
Z
C
¯ z dz where C is the line segment from−1 − i to 3 + i.
(3) [5%] Compute
Z
C
z+ i z3+ 2z2 dz,
where C is the circle|z + 2 − i| = 2 traverse once counterclockwise.
7. The Legendre polynomial Pn(z) is defined by Pn(z) = 1 2nn!
dn dzn
h
(z2− 1)ni . (1) [5%] Use Cauchy’s formula to show that
Pn(z) = 1 2πi
Z
Γ
(ζ2− 1)n 2n(ζ − z)n+1 dζ.
whereΓ is simple closed positively oriented contour containing z in its interior.
(2) [5%] Taking C to be the circle with center at z and radiusp|z2− 1|, show that Pn(z) = 1
π Z π
0
(z +p
z2− 1 cos θ)ndθ, which is known as Laplace’s formula.
(3) [5%] Using Laplace’s formula, show that P2(z) = 12(3z2− 1).
8. [10%] Let f be analytic in the simply connected domain D and let z1and z2be two complex numbers that lie interior to the simple closed contourΓ having positive orientation that lies in D. Show that
f(z2) − f(z1) z2− z1
= 1
2πi Z
Γ
f(z)
(z − z1)(z − z2)dz.
State what happen when z2 → z1.
9. [10%] Let f be a non-constant analytic function in the closed disk|z| ≤ 1. Suppose that |f(z)| is constant on the circle|z| = 1. Show that f has a zero in the domain |z| < 1.