MATH 120A
Your name:
Professor: Elham Izadi
February 9 2017: Solutions to First midterm
(1) (12 points)
(a) Give the definition of the boundary of a subset S of R2 .
(b) Define what it means for a function to be analytic at a point.
Solution:
(a) The boundary of S is the set of all points z such that every ✏-neighborhood of z contains
points in S and points not in S.
(b) A function f is analytic at a point z0 if it is di↵erentiable on some ✏-neighborhood of
z0 .
(2) (22 points) Prove that if f (z) and f (z) are both analytic on a domain D, then f is constant
on D.
Solution: If f and f are analytic on D, then they satisfy the Cauchy-Riemann equations
on D. Writing f (z) = f (x, y) = u(x, y) + iv(x, y), we have f (z) = u(x, y) − iv(x, y).
The Cauchy-Riemann equations for f and f are:
It follows that
ux = vy , uy = −vx , ux = (−v)y , uy = (−v)x
vy = −vy ⇒ vy = 0 ⇒ ux = 0
and
So f ′ = ux + ivx = 0 on D.
vx = −vx ⇒ vx = 0 ⇒ uy = 0.
We proved that if the derivative of an analytic function is zero on a domain then that
function is constant on that domain.
So f is constant on D.
(3) (24 points)
(a) Compute (1 − i)13 . Give your answer in the form a + bi. Simplify your answer as much
as possible.
√
(b) Compute the moduli and the arguments of all the 6-th roots of −32 2(1 − i).
Solution:
(a) We first write (1 − i) in polar form:
1−i=
√
√
1
i
2 � √ − √ � = 2e−i⇡�4 .
2
2
√
√ 13
√
√
13
1
1
(1 − i)13 = � 2e−i⇡�4 � = 2 e−13i⇡�4 = 64 2e3⇡�4 = 64 2 �− √ + i √ � = −64 + 64i.
2
2
(b) We have
√
1
i
−32 2(1 − i) = 64 �− √ + √ � = 64e3i⇡�4
2
2
√
The 6-th roots are the six complex numbers with modulus 6 64 = 2 and arguments
1
6 (3⇡�4 + 2k⇡)
where k takes the values 0, 1, 2, 3, 4, 5.
(4) (14 points) Determine whether the following 3 subsets of C are open, closed, bounded or
neither.
(a) 0 < Im (i + 1)z < 1, −1 < Re z < 3
(b) −2 ≤ Im z ≤ 1, �z + 3 − 2i� ≤ 3
(c) −2 < Im z ≤ 1, �z + 3 − 2i� ≤ 3
Solution:
(a) 0 < Im (i + 1)z < 1, −1 < Re z < 3 Here we have
So the set is
(i + 1)z = (i + 1)(x + iy) = (x − y) + i(x + y)
0 < x + y < 1, −1 < x < 3
this is an open parallelogram. So it is open and bounded.
(b) −2 ≤ Im z ≤ 1, �z + 3 − 2i� ≤ 3 This is the part of the closed disc of radius 3 and center
2i − 3 which lies below the line y = 1. It is closed and bounded.
(c) −2 < Im z ≤ 1, �z + 3 − 2i� ≤ 3 This is the same as the previous set.
(5) (28 points) Consider the function
z 4
f (z) = � �
z
(a) What is the domain of definition of f ?
(b) What is the image of f ?
(c) Does the limit of f exist when z → 0?
Solution:
(a) f is well-defined when z ≠ 0.
(b) Writing z in polar coordinates z = rei✓ , we have
4
z 4
Rei✓
f (z) = � � = � −i✓ � = e8i✓ .
z
Re
This belongs to the unit circle.
As ✓ takes all real values, f (z) takes all values on the unit circle.
So the image of the plane is the unit circle.
(c) z → 0 when r → 0.
Taking z = r on the positive real ray, we obtain f (z) = 1 and limr→0 f (r) = 1.
Taking z = rei⇡�8 , we have f (z) = −1 and limr→0 f (rei⇡�8 ) = −1.
Since f has di↵erent limits along these two rays, f (z) does not have a limit as z → 0.