Name: Solutions
1. (4 points each) Let z =
3
2
√ −1 + i 3 and w = 2 + 3i.
(a) Calculate w−1 .
We have
w−1 =
w
ww
2 − 3i
(2 + 3i)(2 − 3i)
2
3
=
− i.
13 13
=
z
(b) Calculate
w
We have
√ !
3 3 3
2
3
− +
i
− i
2
2
13 13
√
√
9 3−6 9+6 3
+
i.
=
26
13
z
=
w
(c) What is |z|?
We have
√ !2
3 3
2
v
u 2
u
3
|z| = t −
+
2
= 3.
(d) What is arg(z)?
If z = reiθ , then we have θ can be given by cos(θ) = − 32 /|z|. Thus, cos(θ) = − 12 . This gives
θ = 2π
3 . Finally, we have that the argument is then given by
arg(z) =
2π
+ 2πn
3
for all n ∈ Z.
(e) Give z in polar form.
We can now read this off from the above information:
z = 3ei
2π
3
.
2
Name: Solutions
2. (10 points each) (a) Calculate log(2i).
We write 2i in polar form as
π
2i = 2ei 2 .
Thus,
log(2i) = ln(2) + i(
π
+ 2πn)
2
for all n ∈ Z.
(b) Calculate (2i)2+i if we define the power function via the branch of the logarithm given by
w = Log−π (z).
First, we observe that Log−π (2i) = ln(2) + i π2 . The definition of the power function now gives
(2i)2+i = exp((2 + i) Log−π (2i))
π = exp (2 + i) ln(2) + i
2
π
= exp ln(4) −
+ i(ln(2) + π)
2
− π2 i(ln(2)+π)
= 4e e
.
3. (15 points) Give all the roots of the polynomial f (z) = z 5 − 3. Plot these roots in the complex
plane.
Write z = reiθ . Then we want to solve r5 e5iθ = 3. Thus, we have r = 31/5 and 5θ = 0 + 2πin for
n ∈ Z. Thus, θ = 2πin
5 . The distinct values that are roots of the polynomial f are thus given by
2πni
1/5
ζn = 3 e 5 for n = 0, 1, 2, 3, 4.
3
Name: Solutions
4. (10 points each) (a) Evaluate the integral
We see that the function f (z) =
Z
C
1
z 2 −1
R
1
C z 2 −1 dz
on the contour given below:
is analytic between the contours pictured above, so we have
1
dz =
2
z −1
Z
C1
1
dz +
2
z −1
Z
C2
z2
1
dz.
−1
We can now use Cauchy’s integral formula to evaluate each integral. For C1 , we have the function
1
g(z) = z+1
is analytic on and interior to C1 , so we have
Z
C1
1
dz =
2
z −1
Z
C1
g(z)
dz
z−1
= 2πig(1)
= πi.
Similarly, for the contour C2 we have h(z) =
Z
C2
1
z−1
1
dz =
2
z −1
is analytic on and interior to C2 so we have
Z
C2
h(z)
dz
z+1
= 2πih(−1)
= −πi.
Thus, the total integral is 0.
4
Name: Solutions
(b) Evaluate the integral
1
C z 2 −1 dz
R
on the contour given below:
In this case the curve C deforms to the curves C1 and C2 pictured. We have
Z
Z
Z
1
1
1
dz =
dz +
dz.
2−1
2−1
2−1
z
z
z
C1
C2
C
We again use Cauchy’s integral formula. The difference is that C2 is negatively oriented, so the
formula applies to the contour −C2 instead. We have with g(z) and h(z) as in part (a):
Z
Z
1
g(z)
dz =
dz
2
C1 z − 1
C1 z − 1
= 2πig(1)
= πi.
and
Z
C2
1
dz = −
2
z −1
Z
−C2
Z
=−
−C2
1
dz
−1
h(z)
dz
z+1
z2
= −2πih(−1)
= πi.
Thus, the total integral around C is 2πi.
5
Name: Solutions
5. (10 + 5 points each) (a) Find a linear fractional transformation that maps the disc |z − 2| < 2
onto the upper half plane h = {z = x + iy : y > 0} and takes 0 to ∞.
Let our linear fractional transformation be given by T (z) = az+b
cz+d with a, b, c, d ∈ C. We first
determine a, b, c, d so that T maps the boundary of the disc to the boundary of the upper halfplane. To do this, we pick three points on the circle and map them to the real line. We already
know we need T (0) = ∞. This gives that we must have d = 0. Thus, T (z) = az+b
cz . Now we just
pick convenient points on the circle. Let T (4) = 1/4 and T (2 − 2i) = 0. If we pick a = 1 then we
have
2 − 2i + b
T (2 − 2i) =
= 0,
4c
. Finally, using that T (4) = 1/4, we
i.e., b = −2 + 2i. We now plug this in to get T (z) = z−2+2i
cz
get
4 − 2 + 2i
= 1/4,
T (4) =
4c
i.e., c = 2 + 2i. Thus, set T (z) = z−2+2i
(2+2i)z . Since linear fractional transformations take clircles to
clircles, this gives that T maps the circle |z − 2| = 2 to the real axis. To see it maps the interior
of the circle to the upper half-plane, we just need to pick a point in the the circle and see where
it maps. Pick z = 2 and observe
2i
2(2 + 2i)
i
=
2 + 2i
1 1
= + i,
4 4
T (2) =
which is in the upper half-plane. Thus, T is a transformation that works.
(b) Show that your linear fractional transformation maps the crescent shaped region inside the
disc |z − 2| < 2 and outside the disc |z − 1| < 1 onto a horizontal strip.
The only thing we need to show to see this is that the transformation takes the circle |z −1| = 1 to
a horizontal line. Note that since T takes clircles to clircles and we already know that T (0) = ∞,
we know that the image of the circle under T is a line. We just need to show it is horizontal. We
pick two points, say z = 2 and z = 1 + i. We already know that T (2) = 14 + 14 i. We calculate
3 1
T (1 + i) = − + i.
4 4
Since the two values have the same imaginary part, they are on the same horizontal line. Thus,
the crescent region is mapped to the horizontal strip {z : 0 < Im(z) < 41 }.
6
Name: Solutions
6. (5 points each) (a) Let f (z) = u(x, y) + iv(x, y). State sufficient conditions on u(x, y) and
v(x, y) to guarantee f is differentiable at the point z0 .
The first order partial derivatives of u and v with respect to x and y must exist everywhere in a
neighborhood of z0 , be continuous at the point z0 , and satisfy ux = vy , uy = −vx at the point z0 .
(b) Show that f (z) =
y+ix
x2 +y 2
is differentiable for all z 6= 0.
y
Here we have u(x, y) = x2 +y
2 and v(x, y) =
We calculate the partial derivatives:
x
.
x2 +y 2
Note the partial derivatives exist for all z 6= 0.
−2xy
+ y 2 )2
x2 − y 2
uy (x, y) = 2
(x + y 2 )2
y 2 − x2
vx (x, y) = 2
(x + y 2 )2
−2xy
.
vy (x, y) = 2
(x + y 2 )2
ux (x, y) =
(x2
We see that for any z0 6= 0, these satisfy the conditions listed in part (a). Thus, f is differentiable
away from 0.