MATH 3160: APPLIED COMPLEX VARIABLES
FINAL EXAM (VERSION C)
1. Find all solutions z to the equation
tanh z = 1 .
Solution:
There are none.
2. (a) Calculate the Laurent expansion about z = 0 of
1
2
z sin
z2
up to 1/z 4 terms.
(b) Calculate
2
Res z sin
z=0
1
z2
Solution:
(a)
2
z sin
1
z2
=1−
1
+ ···
6z 4
(b)
2
Res z sin
z=0
3. Calculate the integral
Z
C
1
z2
=0
cos z
dz
z (z 2 + 8)
where C is the square with corners at the four points ±2 ± 2i.
Solution: Only one of the poles is inside the square, so
Z
cos z
cos z
π
dz
=
2πi
Res
=
i.
2
z=0 z (z 2 + 8)
4
C z (z + 8)
Date: May 6, 2002.
1
2 MATH 3160: APPLIED COMPLEX VARIABLES FINAL EXAM (VERSION C)
4. Calculate the integral
Z
|z|=1
cos (1/z) dz
.
z 2 sin (1/z)
Hint: why can’t you just use residues to get it? What happens
if you use the coordinate change w = 1/z?
Solution:
There are infinitely many poles inside the circle,
and none outside. Let w = 1/z. Calculate residue at infinity.
Z
Z
cos (1/z) dz
cos(w)
=
dw
2
|w|=1 sin(w)
|z|=1 z sin (1/z)
= 2πi Res
w=0
cos(w)
sin(w)
= 2πi.
5. Show that the map
w=
z−1
z+1
1/2
(with the principal branch of the square root) takes the z plane
slit along the segment −1 ≤ x ≤ 1 onto the right half plane
u > 0 (where z = x+iy and w = u+iv). Hint: it is a composite
function. Figure out where the linear fractional transformation
Z = (z − 1)/(z + 1) takes the upper and lower half planes, and
where it maps the segment −1 ≤ x ≤ 1.
Solution:
As in figure 5 on the facing page, the map Z =
(z − 1)/(z + 1) takes the interval of the real line from z = −1
to z = 1 into the interval of the real line from Z = −∞ to
Z = 0. It takes z = ∞ to Z = 1. It takes real numbers to
real numbers, and nonreal numbers to nonreal numbers, since
we can easily calculate z = −(Z + 1)/(Z − 1). But it takes
z = i to Z = (i − 1)/(i + 1) = i, so it must take the upper half
plane to the upper half plane, and the lower half plane to the
lower half plane. Finally w = Z 1/2 takes the Z plane slit along
the negative real axis to the right half plane in the w plane,
just dividing angles from the real axis in half. Putting these
together, we find the z plane slit along the real axis from −1 to
1 goes to the right half plane.
6. Using conformal mapping, find the harmonic function f which
is defined outside of the two circles drawn in figure 6 on page 4
(unfortunately they don’t quite look like circles, because of some
MATH 3160: APPLIED COMPLEX VARIABLES FINAL EXAM (VERSION C) 3
z
Z
w
problems with graphics), with f = 0 on the little circle, and
f = 1 on the big circle. (The answer looks like figure 6 on
page 5.) Show that the isotherms and flow lines are circles, and
explain why they are perpendicular to each other.
Solution:
Using w = 1/z to straighten out the circles, the
result is
2
x
f=
+1 .
3 x2 + y 2
The isotherms are constant lines of f , so
(x − c)2 + y 2 = c2
for any constant c while the flow lines are
x2 + (y − c)2 = c2
for any constant c.
7. Suppose that f (z) is an entire analytic function, and that the
real part of f (z), which we will call u(x, y), has an upper bound:
u(x, y) ≤ u0
4 MATH 3160: APPLIED COMPLEX VARIABLES FINAL EXAM (VERSION C)
1
0.5
–1
–0.5
0.5
1
1.5
2
–0.5
–1
for any x, y. Show that f (z) constant. Hint: consider g(z) =
ef (z) .
Solution: Applying Liouville’s theorem to g(z) we find that
|g(z)| = eu
is bounded, so g(z) is constant. Taking logarithm of g(z), f (z)
must also be constant.
MATH 3160: APPLIED COMPLEX VARIABLES FINAL EXAM (VERSION C) 5
1
0.5
0
2
2
1
1
y
0
0
–1
–1
–2
–2
x