Lemma. (Cauchys Estimate) [S&T10.5] Suppose that f is dierentiable in cfw_z
C : |z z0 | < R. If 0 < r < R and |f (z)| M for |z z0 | = r then, for all n 0,
|f (n) (z0 )|
M n!
.
rn
Digression. Let D be a domain. The following estimate is extremely useful.
Alternative notation:
df
= Df (z).
dz
Proposition 1. [S&T4.1] If f is dierentiable at z0 then f is continuous at z0 .
f (z) =
Proof. For f to be continuous, we need show that limzz0 f (z) = f (z0 ), i.e.,
limzz0 (f (z) f (z0 ) = 0.
We have
lim (f (z) f (z
MATH 20142 Complex Analysis
Solution Sheet 4
1. (a), (b) Since cos(z) = cos z, sin(z) = sin z, we have
eiz = cos z + i sin z,
eiz = cos z i sin z,
and (a), (b) follow from adding and subtracting these two identities and subsequent dividing
by two.
(c) Fro
MATH 20142 Complex Analysis
Solution Sheet 5
1. (a) We have z(t) = eit = cos t i sin t
(in the clockwise direction).
( t 0), whence our curve is the semicircle
(b) Here z(t) = x(t) + iy(t), where x(t) = 1 + 2 cos t, y(t) = 1 + 2 sin t, 0 t 2. Hence
(x 1)2
MATH 20142 Complex Analysis
Solution Sheet 3
1. (a) The radius of convergence
n
= 1,
n n 1
R = lim |an1 /an | = lim
n
where the disc of convergence is cfw_z : |z| < 1.
(b) Using i2 = 1, i3 = i, i4 = 1, we obtain
n=1
in
i
1
i
1
= +
n
1 2 3 4
i
1
i
1
+
5 6
MATH 20142 Complex Analysis
Solution Sheet 7
1. Let f (z) =
+
m= cm z
m
denote the Laurent series of f at z = 0.
(a) This function has no singularity at 0, whence its Laurent series is none other than its
Taylor series, i.e.,
(
)
1
1
1
z z2
z3
=
= 1+ +
+
MATH 20142 Complex Analysis
Solution Sheet 8
Let R denote SR + [R, R], where SR (t) = cfw_eit , : 0 t .
1. (a) Notice rst that our integral converges, since
dx
<
2+3
x
dx
< +.
x2
1
1
Put f (z) = 1/(z 2 + 3). This function has two poles - 3i and 3i, but on
MATH 20142 Complex Analysis
Solution Sheet 6
1. For a sketch drawing of this contour see [Stewart & Tall, Fig. 12.1].
Let us compute the integral. We have
R
dt
= ln R ln r,
t
iReit
dt = i,
Reit
f=
1
r
f=
2
0
r
f=
R
0
3
f=
4
whence
dt
r
= [ln(t)]R = ln r l
MATH 20142 Complex Analysis
The University of Manchester, 2013
Solutions
Section A
A1. The triangle inequality: |z + w| |z| + |w| for all z, w C.
[bookwork, 1 mark]
Sample proof:
|z + w|2 = (z + w)(z + w) = (z + w)(z + w)
= zz + ww + zw + wz = |z|2 + |w|2
MATH 20142 Complex Analysis
Solution Sheet 2
1. (a) For each z0 C we have
2
z 2 + z (z0 + z0 )
zz0
z z0
(z z0 )(z + z0 + 1)
= lim
zz0
z z0
= lim (z + z0 + 1) = 2z0 + 1,
f (z0 ) = lim
zz0
whence f (z) = 2z + 1.
(b) For z0 = 0,
2
1/z 2 1/z0
zz0
z z0
2
z0 z
MATH 20142 Complex Analysis
The University of Manchester, 2011
Solutions
Section A
A1. (a) A function f is called continuous at z0 if limzz0 f (z) = f (z0 ).
[bookwork, 1 mark]
(b) A function f is called dierentiable at z0 if there exists the limit
lim
zz
MATH 20142 Complex Analysis
The University of Manchester, 2012
Solutions
Section A
A1. A function f is called dierentiable at z0 if there exists the limit
lim
zz0
f (z) f (z0 )
.
z z0
[bookwork, 2 marks]
By denition,
z3 z2 + z 1
= lim (z 2 + 1) = 3.
z1
z1
Example.
n=0
an z n =
n=0
z n /n2 . Then
lim |an1 /an | = lim n2 /(n 1)2 = 1,
n
n
so the radius of convergence is 1. For |z| = 1 we have |z n /n2 | = 1/n2 , whence the
series converges for all z with |z| = 1.
More generally:
Theorem. [S&T3.8] The radius o
4. Power Series
Sequences and series. We say a sequence sn C converges to s C if, given
any > 0, there exists N N such that |sn s| < for all n N .
The series k=0 zk (we dont need to start at k = 0) converges if the sequence
n
of partial sums sn = k=0 zk c
Innite Real Integrals. Now we are going to look at some integrals of the form
f (x) dx.
At rst sight, it is not clear what this has to do with complex integrals.
First we have to be clear about what the innite integral means. Formally, we
say that f (x) d
Evaluating integrals. The most important use of the Residue Theorem is as a
tool for evaluating integrals. It turns out that often this can give an easy way of
evaluating real denite integrals which would be hard by conventional (real calculus)
methods. T
9. Residues and evaluation of integrals
[This corresponds to Chapter 12 of S&T.]
Cauchys Residue Theorem. If f has an isolated singularity at z0 and Laurent
expansion
f (z) =
an (z z0 ) +
n
n=0
bn (z z0 )n ,
0 < |z z0 | < R,
n=1
we dene the residue of f a
8. Laurent series
Theorem (Laurents Theorem). [S&T11.1] If f is dierentiable in the annulus
cfw_z C : R1 |z z0 | R2 ,
where 0 R1 < R2 , then
f (z) =
an (z z0 ) +
n
n=0
bn (z z0 )n ,
n=1
where n=0 an (zz0 )n converges for |zz0 | < R2 and n=1 bn (zz0 )n con
Recall: if f is dierentiable in a simply connected domain D, then
any closed contour D.
f = 0 for
Example. Let D = C, f (z) = z 2 and
(t) = cfw_eit : 0 t + [1, 1].
Then
2
f = 3 +
2
3
= 0.
Example. Let D = C \ cfw_0, f (z) = 1/z and (t) = cfw_eit : 0 t 2.
The logarithmic function.
Let z C. Consider
exp w = z.
()
By above, if w1 is a solution of (*) then so is w1 + 2ni. Each of these values is
called a logarithm of z and is written log z.
If x R, x > 0 then
exp w = x
has a unique real solution. We call this
Theorem 6.1 (Fundamental Theorem of Contour Integration). [S&T6.7]
If f : D C is continuous, F : D C satises F = f and is a contour in D
from z0 to z1 then
f = F (z1 ) F (z0 ).
(More precisely: : [a, b] D with (a) = z0 , (b) = z1 .)
Proof. Suppose w(t) =
6. Integration
Paths. A path is a function : [a, b] C, where [a, b] is a real interval. So, for
a t b, (t) denotes a point on the path.
Example. For z1 , z2 C, : [0, 1] C given by
(t) = (1 t)z1 + tz2
is the straight line between z1 and z2 . (Picture.)
Not
5. The Exponential Function and its Friends
The exponential function. Dene
exp z =
zn
.
n!
n=0
This has radius of convergence R where
1/R = lim sup(1/n!)1/n = 0,
n
so R = and the series converges absolutely for all z C.
We may dierentiate term by term (T
However there is a partial converse.
Theorem 3.4. [S&T4.6] If f (z) = u(x, y) + iv(x, y) is a complex function on an
open set S and at z0 = x0 + iy0 S the partial derivatives
u v u v
,
,
,
x x y y
all exist, are continuous and satisfy the C-R equations
v
Topology in C
Limits, open and closed sets. Earlier, we wrote a limit; we shall begin by
making clear what taking limits in C means.
Notation. We shall use (not ) to denote is a subset of.
Denition. For z0 C and
> 0, we shall write
N (z0 ) = cfw_z C : |z
In consequence
ez = ex+iy = ex eiy = ex (cos y + i sin y),
|e | = e | cos y + i sin y| = e
cos2 y + sin2 y = ex .
and
z
x
x
Corollary.
(1)
(2)
(3)
(4)
(5)
cos z = 1 (eiz + eiz )
2
1
sin z = 2i (eiz eiz )
cos2 z + sin2 z = 1
sin(z + w) = sin z cos w + cos
MATH 20142 Complex Analysis: Test
Manchester, March 14, 2014
Answer ALL four questions
Problem 1
Find all complex solutions of the following quadratic equation:
z(z + 2) = 1 + 18i
Solution. We have
(z + 1)2 = 18i.
Denote z + 1 = a + ib; then a2 + 2iab b2