Math 228
Assignment 1 - Solutions
[2]
1. By multiplying top and bottom by the complex conjugate of the denominator, we get:
(a)
(b)
[2]
(1+i)(23i)(3i)
32 +12
(5i)(3i)
= 7 4 i.
10
5
5
2i(8+6i)2
2i(2896i)(32+i(20 5)
2i(8+6i)2 (4+2 5i)(2+4 5i)
=
=
(16+45)(4+

Math 228
Assignment 9 - Solutions
[1]
1. (a) (= 4.6.10) Since |f (0)| = |i| = 1, and since |f (z)| 1, for all z D, we see that |f | attains its maximum
at z = 0. By a class theorem (Theorem 23 in the text), it follows that f is constant. Thus f (z) = f (0

Math 228
Assignment 4 - Solutions
[3]
1. (a) (= 3.1.3b) Clearly, z 4 16 = (z 2 4)(z 2 + 4) = (z 2)(z + 2)(z 2i)(z + 2i), which is the complete
factorization of z 4 16.
7
1
(b) (= 3.1.3c) We have (z 1)(1 + z + . . . + z 6 ) = z 7 1, so 1 + z + . . . + z 6

Math 228
Assignment 3 - Solutions
[1]
g(z)
1. (a) (= Ex. 2.3.7d) Here f (z) = h(z) with g(z) = (z + 2)3 and h(z) = (z 2 + iz + 1)4 . Now by the chain rule
we have g (z) = 3(z + 2)2 and h (z) = 4(z 2 + iz + 1)3 (2z + i), so by the quotient rule we obtain
f

Math 228
Assignment 2 - Solutions
[1]
1. (a) (= Ex. 1.6.4d) The set S1 = cfw_z C : 1 < Imz 1 is not open and hence is also not a domain.
r
Indeed, i S1 (because Im(i) = 1) but no open disk D(i, r) is contained in S1 because (1 + 2 )i D(i, r)
r
but (1 + 2

Math 228
Assignment 5 - Solutions
[1]
1. (a) (= 3.2.1) Put z0 = i . Then ez0 = cis( ) =
4
4
ez =
1+i
2
ez = ez0
1+i
.
2
Thus,
z = z0 + 2ik
for some k Z,
the later by Theorem 3(b) in class. This proves the assertion because z0 + 2ik = i( + 2k).
4
[2]
(b)

Math 228
Assignment 6 - Solutions
[1]
1. We rst note that arg (z) = Arg(z) + 2, for all z = 0. Indeed, since arg (z) = Arg(z) + 2k, for some
integer k, and since Arg(z) (, ], we see that Arg(z) + 2 (, 3], so arg (z) = Arg(z) + 2.
Thus, by denition L (z) =

Page
I
of 7
Student Name:
Student Number:
MATH/MTHE
228
- COMPLEX ANALYSIS
MIDTERM EXAMII\TATIOIcfw_
MARCH 8,2AL2
INSTRUCTOR: M. BRANNAN
INSTRUCTIONS
o This exam 90 minutes long. No personal aids other than approved calculators
are
allowed.
o Answer all q

Math 228
Assignment 11 - Solutions
[1]
1. (a) (= 5.6.1b) Since z 3 is analytic everywhere and e1/z everywhere except at 0, where it has an essential
singularity, we see that 0 is the only isolated singularity of f (z) = z 3 e1/z , and that it is an essent

Math 228
Assignment 10 - Solutions
[2]
1. (a) (= 5.2.4) Put f (z) = eLog(1+z) . Then I claim that
f (k) = f (z)
(1)
( 1) . . . ( k + 1)
,
(1 + z)k
for all k 1.
This is easily veried by induction. Indeed, by the chain rule we obtain that f (z) = f (z)(Log(

Math 228
Assignment 8 - Solutions
[1]
1. (a) (= 4.3.1(c) By chapter 3 we know that the principal branch Log(z) is an anti-derivative of
D = C \ (, 0]. Since any one of the given contours lies completely in D , we thus obtain
[1]
in
i
i
dz
= Log(3i)Log(3i)

Name:
Student Number:
Queens University
Faculty of Arts and Science
Department of Mathematics and Statistics
Math/Mthe 228 Midterm Test
7 March 2013
12:3013:20
Professor E. Kani
Instructions:
Write your name and student number
at the top of this page. Th