MATH 208, SPRING, 2013, PRACTICE FINAL EXAM SOLUTIONS
(1) State the Cauchy Integral Formula, being careful to include the conditions on the function
f (z). (10 points)
Solution: Let f be analytic on and inside a simple closed contour C , positively orient
Math 208
Elementary Functions
January 27, 2013
Exponentials
Already discussed.
Logarithms
Its quite easy, at this stage, to dene the logarithm, log(z) := w if
e w = z. The only diculty with that denition is that the function is not
well-dened. That is, th
Math 208
Integration
February 4, 2013
Complex-valued functions of t
We will be dealing with, from now on, complex-valued functions of a real
variable t, in one shape or another, w (t) = u(t) + iv (t). We can clearly
apply ordinary calculus formulas to the
Math 208
Laurent Series and Power Series
March 8, 2013
Laurents Theorem
Theorem
Suppose that f is analytic on an annular domain R1 < |z z0 | < R2 , and
let C be any simple closed curve around z0 inside the annulus. Then,
n
an (z z0 ) +
f (z) =
n=0
n=1
bn
Math 208
Series
February 14, 2013
Sequences
Denition
A sequence of complex numbers (or anything else) is, technically just a
function from the positive integers to C. We usually call the positive
integers the natural numbers, N, so a sequence is a map : N
MATH 208, SPRING, 2013, EXAM # 1 SOLUTIONS
(1) Find all points z C so that z 6 = 1. (10 points)
Solution: If z = rei in polar form, then z 6 = 1 is equivalent to r6 ei6 = 1, so
r = 1 and 6 = 2n for some n, or = n . To stay within the principal branch
3
of
MATH 208, SPRING, 2013, FINAL EXAM SOLUTIONS
(1) State the Cauchy-Goursat theorem.
(10 points).
If a function f is analytic at all points of a domain
and its interior, then C f (z)dz = 0.
Theorem.
contour
C
D containing a simple closed
(2) State Taylor's
SOLUTIONS TO THE MATH 208, SPRING, 2013, REVIEW SHEET FOR
EXAM # 2
ez
dz , where C is the unit circle |z| = 1, with the usual
(1) Find the contour integral
2
C z
orientation. Use the corollary to the CIF.
Apply, as I said, that powerful corollary to the C
MATH 208, SPRING, 2013, EXAM # 2 SOLUTIONS
(1) State Liouvilles theorem, being careful to include the conditions on the function f (z).
(10 points)
Solution: A bounded, entire function is constant.
(2) Let f (z) be an entire function so that there is a po
SAMPLE PROBLEMS FOR THE FINAL
The sample problems should not be taken as a complete list problems that you will
be tested on in the nals and they are just to provide you problems for practice.
Students are encouraged to work additional problems from the t
SOLUTIONS FOR THE SAMPLE PROBLEMS FOR THE FINAL
The sample problems should not be taken as a complete list problems that you will
be tested on in the nals and they are just to provide you problems for practice.
Students are encouraged to work additional p
MATH 208, SPRING, 2013, PRACTICE EXAM # 1 SOLUTIONS
(1) Find all points z C so that z 3 = 8.
Us the polar form of z = rei , where we will use the principal branch of the
argument, < . If z 3 = 8, then r3 ei3 = 8, so 3 = 2n , or = 2n and
3
r = 2. Within th
Math 208
Complex Numbers
January 12, 2013
Why Complex numbers?
The basic idea of complex numbers is to make things simpler. For
instance, in terms of real numbers the polynomial equation x 2 1 = 0 has
two solutions, x = 1 and x = 1, but the equation x 2 +
Math 208
Derivatives
January 23, 2013
The derivative
There is only one possible denition for the derivative of a function of a
complex variable:
Denition
A complex function w = f (z) dened on a domain D containing z0 is
(z
dierentiable at z0 , with deriva
MATH 208, SUMMER II, 2016
HW #1
Due Monday, 7/11. This material should be found in Chapter 1 of the text.
Show that z = 1 i satisfies the equation z 2 2z + 2 = 0.
2 + z + 1 = 0.
Solve the
z
equation
1
i
.
Find 3i
2+3i
Reduce each of these expressions to a
Math 208, Summer II, 2016
July 10, 2016
HW #2
Due Thursday, 7/14.
Problems from the book
2.1 14, 28, 32
2.3 18, 32
2.4 22
3.1 22
3.2 6
3.3 6, 30
3.4 10
Non-book problems
1. Write f (z) = z +
1
z
in terms of its real and imaginary parts, f (z) = u(x, y) +
MATH 208, SUMMER II, 2016
HW #5
Due Tuesday, 8/9.
(1) Compute the following residues: (10 points each)
1
.
(a) Resz=i f (z), where f (z) =
2
(z + 1)2
(b) Resz=i f (z) and Resz=2i f (z), where f (z) =
1
.
3z 2 + 10z + 3
Evaluate the following real integral
MATH 208, SUMMER II, 2016
Syllabus
Text. Zill and Shanahan, Complex Analysis, a first course with applications, 3rd edition. I plan to
cover essentially the entire text.
Prerequisites. At least three semesters of calculus are absolutely essential, includi
MATH 208, SUMMER II 2016, HW #5 SOLUTIONS
(1) Compute the following residues:
(a)
Resz=i f (z),
where
Solution:
f (z) =
The pole at
(10 points each)
1
(z 2
i
.
+ 1)2
is a pole of order 2, since
Resz=i f (z) =
=
=
z2 + 1
2
= (z + i)2 (z i)2 ,
0
2
f (z) (
MATH 208, SUMMER II, 2016
HW #4
Due Thursday, 7/28.
Problems from the book.
6.1: 14, 20, 38 (the answers are yes, and no, in that order.), 40
1
1
14: This is a trick. Since k(k+1)
= k1 k+1
by partial fractions,
n
X
k=1
1
k(k + 1)
=
1 1
1 2
+
1 1
2 3
+ +
1
MATH 208, SUMMER II, 2016
HW #3 S OLUTIONS
Problems from the book.
4.1: 12, 36, 52,
12:
f (z) = e1/z
= e
= e
x
i 2 y 2
x2 +y 2
x +y
x
x2 +y 2
x
= e x2 +y2
y
y
cos
i sin
x2 + y 2
x2 + y 2
x
y
y
2 +y 2
x
cos
ie
sin
x2 + y 2
x2 + y 2
36: To solve equations
MATH 208, SUMMER II, 2016
HW #1 S OLUTIONS
(1) Show that z = 1 i satisfies the equation z 2 2z + 2 = 0.
Solution:
(1 i)2 2 (1 i) + 2 = (1 2i 1) 2 2i + 2
= 0.
(2) Solve the equation z 2 + z + 1 = 0.
Solution: Use the quadratic formula to find the roots: z
MATH 208, SUMMER II, 2016
HW #4
Due Thursday, 7/28.
Problems from the book.
6.1: 14, 20, 38 (the answers are yes, and no, in that order.), 40
6.2: 20, 26, 32
6.3: 4, 8, 24, 32
6.4: 18, 36
Non-book problems:
2
3z +1
(1) Find the Taylor series, centered at
Applications
March 28, 2013
Improper integrals
Improper integrals are of several sorts, but most of what we will deal with
are integrals of the form f (x ) dx . Of course, convergence of such an
integral is the existence of the double limit
b
lim
b
lim
a
Math 208
Cauchy Integral Formula
February 10, 2013
CIF
Theorem
[Cauchy Integral Formula] Let f be analytic on and inside a simple
closed contour C , positively oriented. If z0 is any point in the interior of
C , then
1
f (z) dz
f (z0 ) =
.
2i C z z0
CIF
T
Math 208
Cauchy-Goursat
February 4, 2013
Antiderivatives
Theorem
The following are equivalent, for a continuous function f on a domain D.
1
f has an antiderivative F in D, that is, an analytic function F so that
F (z) = f (z) for all z D.
2
contour integr