MAT 421 Test 1. Spring 2017. Jennings
SHOW YOUR WORK!
3 + 4i
1. Compute the quotient
in rectangular coordinates (“x+iy”).
5 + 6i
Solution:
3 + 4i
5 + 6i
5 − 6i
5 − 6i
=
(15 + 24) + i(20 − 18)
39 + 2i
=
25 + 36
61
2. Write 2eiπ/6 in rectangular coordinates (“x+iy”).
Solution:
√
2 cos(π/6) + i sin(π/6) = 3 + i
3. Find the product 2eiπ/3 4eiπ/6 . Express your answer in polar coordinates (“ rei/θ ”).
Solution:
8ei(π/3+π/6) = 8eiπ/2
5
4. Find 2(cos(π/3) + i sin(π/3)) . Express your answer in polar coordinates: (“ rei/θ ”).
Solution: Use polar coordinates:
5
5 2(cos(π/3) + i sin(π/3)) = 2eiπ/3 = 25 ei5π/3 .
5. Find all three values of
√
3
1 + i.
√
Solution: In polar coordinates, 1 + i = 2eiπ/4 . The angle is not unique; one can add any integer
multiple of 2π to the angle without changing anything so
√
√
√
√
1 + i = 2eiπ/4 = 2e(iπ/4±i2π) = 2e(iπ/4±i4π) = 2e(iπ/4±i6π) = · · ·
√
= 2ei(π/4+2nπ) where n = 0, ±1, ±2, ±3, . . .
Thus the cube root is
√ 1/3 i(π/4+2nπ) 1/3
(1 + i)1/3 =
2
e
= 21/6 ei(π/12+2nπ/3) where n = 0, ±1, ±2, ±3, . . .
Of all the angles π/12+2nπ/3, n = 0, ±1, ±2, ±3, . . ., only three, π/12, π/12+2π/3, and π/12+4π/3,
actually point in different directions in in the plane. The rest differ from one of those three by
multiples of 2π so they don’t point in any new directions or produce any new complex numbers.
Thus there are exactly three cube roots of 1 + i:
21/6 eiπ/12 ,
21/6 ei(π/12+2π/3) ,
and
21/6 ei(π/12+4π/3)
6. (a) The definition of “complex-derivative” says f is differentiable at z0 if and only if a certain limit
exists. What is that limit?
Solution:
lim
h−>0
f (z0 + h) − f (z0 )
h
There are several equivalent ways to write this limit, for example lim
z→z0
all OK of course.
f (z) − f (z0 )
. They are
z − z0
(b) Suppose f is only defined at points x + iy where x is is least as large as the real part of z0 . (In
other words, where x ≥ Re(z0 )). Is it possible for f to be complex-differentiable at z0 ? Why, or
why not?
Solution: No. The above definition of “differentiable” requires that h be able to approach 0
from any direction. If f (z) is not defined when the real part of z is less that the real part of z0
then f (z0 + h) will not be defined when h approaches 0 from the left-half plane {(x, y) | x < 0}
so f isn’t differentiable at z0 .
7. (a) Let f : C → C be a holomorphic function, f (x + iy) = u(x, y) + iv(x, y) with x, y and u, v real.
What do the Cauchy-Riemann equations say about u and v?
Solution:
∂v
∂u
∂v
∂u
=
and
=− .
∂x
∂y
∂y
∂x
(b) Show that the function g(z) = z̄ is not holomorphic.
Solution: g(x + iy) = x − iy so u(x, y) = x and v(x, y) = −y. Thus
∂u
= 1,
∂x
∂v
= −1
∂y
so the first Cauchy-Riemann equation is not satisfied.
az + b
that
cz + d
(a) maps the unit circle |z| = 1 to the x-axis (including the point at infinity on the x-axis).
8. Find a formula for a linear fractional transformation f (z) =
Solution: It’s enough to find a linear fractional transformation that sends three points on the
unit circle to three points on (x-axis) ∪ {∞}. For example let f send 1 7→ 0, i 7→ 1, −1 7→ ∞:
z−1
i+1
f (z) =
.
z+1
i−1
(b) maps the y axis to the circle |z| = 2.
Solution:
Page 2
It’s enough to find a linear fractional transformation f that sends three points on (y-axis)∪{∞}
to three points on the circle of radius 2. For example
f (0) = 2,
f (i) = 2i,
f (∞) = −2 .
One way to do this would be to simply plug all these numbers into the formula f (z) = az+b
cz=d to
obtain some equations, then solve them for a, b, c, d.
Another way to do this would be to start by finding a formula for the inverse function f −1 ,
which would send three points on the circle of radius 2 to three points on (y-axis) ∪ {∞}.
0 = f −1 (2),
i = f −1 (2i),
∞ = f −1 (−2) .
Then one could invert the formula for f −1 to get a formula for f .
I’ll do it the second way, but the first way works equally well.
z−2
does that, so probably
z+2
by multiplying that formula by some constant:
z−2
−1
f (z) = k
z+2
Second way: with f −1 we need 2 7→ 0 and −2 7→ ∞. The function
we can get a formula for f −1
where k is some constant. To find out what constant to use, plug in the equation i = f −1 (2i):
2i − 2
i=k
= (k)(i)
2i + 2
k=1.
Thus
f −1 (z) =
z−2
.
z+2
Now solve the equation w = f −1 (z) for z to obtain a formula for f .
z−2
z+2
w(z + 2) = z − 2
w = f −1 (z) =
wz + 2w = z − 2
wz − z = −2w − 2
z(w − 1) = −2(w + 1)
w+1
z = −2
= f (w)
w−1
(The answers to parts a and b probably are different.)
Z
9. Compute the line integral
f (z) dz of the function f (z) = z 3 + 2z + 1 around the square C whose
C
vertices are (1, 1), (−1, 1), (−1, −1), (1, −1), oriented counterclockwise.
Hint: this is very easy.
Page 3
Solution: It’s zero because the path is closed and f is a polynomial, so f is holomorphic on the
whole complex plane.
10. Let D be the simply-connected domain obtained by removing the upper -half of the imaginary axis
{iy | y ≥ 0} from the complex plane. Let γ be a smooth path in D that starts at 1 and ends at the point
−1 + i. Compute the line integral
Z
dz
.
γ z
Hint: Sketch a diagram to make sure you understand the problem. You don’t need to parametrize the
path or compute the line integral using real coordinates etc. Instead, approach the problem by asking
yourself: What is the antiderivative of 1/z?
Solution: Any version of the natural logarithm function that is defined on D will do for an antiderivative of 1/z. One can define a natural log on D by restricting angles to the range −3π/2 <
θ < π/2:
3π
π
ln reiθ = ln(r) + iθ if −
<θ< .
2
2
The line integral is the difference of the values of this log function at the endpoints of γ:
Z
√
dz
1
5π
−5/π
= ln(−1 + i) − ln(1) = ln( 2) + i
− 0 = ln(2) − i
.
z
4
2
4
γ
Page 4