MATH 3160: APPLIED COMPLEX VARIABLES
FINAL EXAM (VERSION B)
1. Find all solutions z to the equation
tan z = 1 .
Solution:
z=
π
+ πN
4
for N any integer.
2. (a) Calculate the Laurent expansion about z = 0 of
cos z
sin z
2
up to z terms.
(b) Calculate
cos z
Res
z=0 sin z
Solution:
(a)
cos z
1 z
z3
= − −
+ O z5
sin z
z 3 45
You only need the first two terms.
(b)
cos z
Res
=1
z=0 sin z
3. Calculate the integral
Z ∞
x2
dx .
4
−∞ x + 1
Solution:
The function
z2
f (z) = 4
z +1
has two poles in the upper half plane, at
1
z = √ (±1 + i) .
2
Date: May 2, 2001.
1
2 MATH 3160: APPLIED COMPLEX VARIABLES FINAL EXAM (VERSION B)
The residues at these points are
1
√ (±1 − i) .
4 2
The integral is the product of 2πi with the sum of the residues.
This gives
Z ∞
x2
π
dx = √ .
4
2
−∞ x + 1
4. Consider the integral
Z
|z|=R
√
e−1/z z 2 − 1
dz .
z3
√
where the branch of z 2 − 1 is any branch defined in the whole
plane except on the real line between x = −1 and x = 1, and
the radius R is bigger than 1.
(a) Show that the integral is less than 4πe/R.
(b) Show that the value of the integral with R = 2 is
√
Z
e−1/z z 2 − 1
dz = 0 .
z3
|z|=2
Solution:
(a) The integrand is never larger in magnitude than
√
e1/R R2 + 1
e(R + 1)
2e
<
< 2 .
3
3
R
R
R
This is integrated over a circle of length 2πR, so the integral
is less than
2e
4πe
· 2πR =
.
2
R
R
(b) By the Cauchy–Goursat theorem, since there are no singularities in between the circle |z| = 2 and the circle |z| = R,
the integrals over these circles must be the same. Letting
R → ∞, we find that the integral goes to zero.
5. Find the potential U (x, y) inside the unit circle so that U = 1 on
a three quarter arc of the circle, and U = 0 on the remaining one
quarter, as drawn in figure 1 on the facing page. The function
U (x, y) is drawn in figure 2 on the next page. Hint: you could
use a linear fractional transformation taking 1, i, −1 to 0, ∞, −1.
MATH 3160: APPLIED COMPLEX VARIABLES FINAL EXAM (VERSION B) 3
Figure 1. The potential on the boundary of the unit disk
1
0.8
0.6
0.4
0.2
0
–1
–1
–0.5
–0.5
0x~
y~ 0
0.5
0.5
1
1
Figure 2. The potential inside the unit disk
Solution:
say
Let us use the hint, and try to find such a map,
w=
az + b
.
cz + d
4 MATH 3160: APPLIED COMPLEX VARIABLES FINAL EXAM (VERSION B)
i
0
1
1
0
1
−1
0
Figure 3. How the z disk maps to the w half plane,
and then to the Z half plane
To get z = 1 to w = 0, we will want z − 1 in the numerator. To
get z = i to w = ∞ we will want z − i in the denominator. So
w=a
z−1
.
z−i
We only need the correct scaling factor a to get w = −1 when
z = −1. Plugging these in gives
a=−
1+i
2
so that
1+iz −1
.
2 z−i
See figure 3. Multiplying numerator and denominator by z̄ + i,
we find (after some algebra)
w=−
w =u + iv
2
−1
=
x + y 2 − 2x − 2y + 1 + i x2 + y 2 − 1
2
2|z − i|
We can simplify this to
w =u + iv
−1
=
(x − 1)2 + (y − 1)2 − 1 + i x2 + y 2 − 1
2
2|z − i|
We have already seen the answer to this problem:
U=
1
arctan(v/u)
π
MATH 3160: APPLIED COMPLEX VARIABLES FINAL EXAM (VERSION B) 5
where w = u + iv. In z = x + iy coordinates,
1 − (x2 + y 2 )
1 − (x − 1)2 + (y − 1)2
1
U (x, y) = arctan
π
6. Calculate the integral
Z
0
Solution:
2π
!
dθ
.
2 + cos θ
Plugging in z = eiθ the integral becomes
Z
dz
−2i
2
|z|=1 z + 4z + 1
and we factor the denominator to find
Z
dz
√ √ .
−2i
3
z − −2 + 3
|z|=1 z − −2 −
√
Only the root z = −2 + 3 lies inside the unit disk. Therefore
the integral is
2π
1
√ √ =
.
−2i · 2πi Res√
3
z=−2+ 3 z − −2 −
3
z − −2 + 3