MA 412 Complex Analysis

Final Exam

Summer II Session, August 9, 2001.

1. Find all the values of (−8i)1/3. Sketch the solutions.1 pt.

Answer: We start by writing −8i in polar form and then we’ll compute thecubic root:

(−8i)1/3 = (8e−iπ/2)1/3

= 81/3 exp(i(−π

6+

2πk

3)),

Hence z0 = 2 exp(−iπ/6), z1 = 2 exp(iπ/2) = 2i, z2 = 2 exp(i7π/6).

2. Suppose v is harmonic conjugated to u, and u is harmonic conjugated to v.Show that u and v must be constant functions.1 pt.

Answer By definition, v is harmonic conjugated to u if Δu = Δv = 0 andux = vy, uy = −vx hold. On the other hand, as u is harmonic conjugated to v,we also have vx = uy and vy = −ux. Then,

uy = −vx = vx implies vx = 0 and uy = 0

andux = vy = −vy implies vy = 0 and ux = 0.

Hence, u(x, y) and v(x, y) are constant functions.

3. Show that f(z) = |z|2 is differentiable at the point z0 = 0 but not at anyother point.1 pt.

1

z 2z 0

z 1

2

= 2i

Figure 1: Roots of (−8i)1/3.

Answer: We could show that f is not differentiable at any z0 ∈ C − 0 bycomputing different limits for different trajectories. But we’ll use the necessaryconditions of differentiability (i.e. the Cauchy-Riemann equations) to do so.

First, we write f(z) = u(x, y)+ iv(x, y), so u(x, y) = x2 +y2 and v(x, y) = 0.Clearly, the first-order partials of u and v are continuous everywhere. But

ux = 2x equals vy = 0, and

uy = 2y equals − vx = 0

if and only if x = y = 0. Then f can only be differentiable at z0 = 0.

4. Construct a branch of f(z) = log(z +4) that is analytic at z = −5 and takeson the value 7πi there. Sketch the branch cut in the complex plane.1 pt.

Answer: Let α = 6π. Then define the branch of log as

logα(z + 4) = ln|z + 4| + iθ

where 6π < θ ≤ 8π. Thus

logα(−5 + 4) = logα(−1) = ln| − 1| + i(π + 6π)

that is logα(−5 + 4) = 7πi.5. Find ∫

C

exp(z)zn

dz

2

α = 6π

7π

−4

Figure 2: The branch cut of log6π.

where C : z(t) = e2πit, 0 ≤ t ≤ 1 and n is any non-negative integer.1 pt.

Answer: Let f(z) = exp(z). Since z0 = 0 lies inside the given contour and f isanalytic in the interior of C, then by Cauchy’s integral formula for derivatives

∫C

exp(z)zn

dz =2πif(z0)(n − 1)!

=2πi

(n − 1)!

for n = 1, 2, . . .. For n = 0, the integrand reduces to f(z) which is an entirefunction. Hence, by Cauchy-Goursat theorem, we conclude

∫C

exp(z)zn

dz = 0 n = 0.

6. Find two Laurent series expansions for

f(z) =1

z3 − z4

that involves powers of z. Use the regions 0 < |z| < 1 and |z| > 1.1 pt.

Answer: The given function has two singularities at z0 = 0 and z1 = 1. We’llproceed by expanding f in the regions, D0 : 0 < |z| < 1 and D1 : |z| > 1.

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At D0,

f(z) =1z3

11 − z

=1z3

∞∑n=0

zn,

for 0 < |z| < 1. Then

f(z) =∞∑

n=0

zn−3

=1z3

+1z2

+1z

+∞∑

n=0

zn,

valid at D0.At D1, |z| > 1 implies 1/|z| < 1. Then we must write f(z) as

1z3

11 − z

=1z4

−11 − 1

z

.

Hence,

f(z) =−1z4

∞∑n=0

1zn

= −∞∑

n=4

1zn

,

valid at D1.

7. Show

1.1

1 + z=

∞∑n=0

(−1)nzn, |z| < 1,

2.1z

=∞∑

n=0

(−1)n(z − 1)n, |z − 1| < 1.

1 pt.

Answers:(a) Using the Taylor series for 1/(1 − z) =

∑∞n=0 zn, valid at |z| < 1, we

obtain

11 + z

=1

1 − (−z)=

∞∑n=0

(−1)nzn

valid also at |z| < 1.

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(b) In this case, we’ll use the result from question (a):

1z

=1

1 + (z − 1)=

∞∑n=0

(−1)n(z − 1)n,

valid for |z − 1| < 1.

8. Use Cauchy’s Residue Theorem to evaluate the next integral around thepositively oriented contour C : |z| = 3

∫C

11 + z2

dz.

1 pt.

Answer Let p(z) = 1 and q(z) = 1 + z2, so f(z) = p(z)/q(z) represents theintegrand given. Notice that q(z) has two zeros at z0 = i and z1 = −i, andboth zeros are simple since q′(z) = 2z = 0 if and only if z = 0. Then f has twosimple poles at z0 and z1, as p, being a constant function, has no zeros at all.By Cauchy’s residue theorem, we obtain

∫C

11 + z2

dz = 2πi(Resz=z0

11 + z2

+ Resz=z1

11 + z2

)

= 2πi( 1

2i+

1−2i

)

= 0

after evaluating q′(z) at z0 and z1.

9. State and prove Liouville’s theorem.1pt.

Answer: Let f be an entire function. If f is bounded, then f reduces to aconstant function.

Assume there exists M > 0 such that |f | ≤ M in the whole complex plane.Fix any point z0 ∈ C. For any R > 0, Cauchy’s inequality states

|f ′(z0)| ≤ M

R

when we restric f to the circular contour CR : |z − z0| = R. As R tends to ∞,the absolute value of f ′(z0) becomes zero. That implies

ux = vx = 0 and uy = vy = 0,

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at (x0, y0). Since z0 was arbitrary, it follows that u and v are constant functionsin the whole complex plane, and f reduces to a constant.

10. Compute ∫ ∞

0

dx

x4 + 1.

1 pt.

Let p(x) = 1 and q(x) = x4 + 1. Clearly, q has complex roots, so q(x) �= 0for all x ∈ R. Since the degree of q is larger than 2, we can apply the theoremof indefinite integrals. We first compute the singularities of the integrand incomplex form. The zeros of q(z) = z4 + 1 are given by the fourth roots of −1,

zk = exp(i(

π

4+

2πk

4))

for k = 0, 1, 2, 3. Only the roots z0 = (1 + i)/√

2, z1 = (−1 + i)/√

2 lie in theupper half plane, so we shall only compute the residues for these two singulari-ties. Also notice that, since q′(z) = 4z3 = 0 if and only if z = 0, we only havesimple zeros. It follows that

Resz=zk

p(z)q(z)

=1

4z3k

.

Hence ∫ ∞

−∞

p(x)q(x)

dx = 2iπ(e

−3iπ4 + e

−9iπ4

)

=iπ

2

(−1√2(1 + i) +

1√2(1 − i)

)

=iπ

2√

2(−1 − i + 1 − i)

=π√2

Since the give integrand is an even function, we conclude∫ ∞

0

dx

x4 + 1=

12

∫ ∞

−∞

dx

x4 + 1=

π

2√

2.

A final question

What is the relationship between Complex Analysis and the animated gif of asolar flare found in the course webpage?

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