M361 (56650) Midterm 2 Solutions
1. (a) (5 points) Let f (z ) = (1 cos z )5 . Find ord0 f .
Solution: We compute
ord0 f = 5 ord0 (1 cos z ) = 10
since 1 cos z = z 2 /2 + O(z 4 ).
(b) (5 points) Let g (z ) = sec z . Find resg /2.
Solution: We have
1
z /2
=
M361 (56650) Midterm 2 Solutions
1. (a) (5 points) Let f (z ) = (1 cos z )5 . Find ord0 f .
Solution: We compute
ord0 f = 5 ord0 (1 cos z ) = 10
since 1 cos z = z 2 /2 + O(z 4 ).
(b) (5 points) Let g (z ) = sec z . Find resg /2.
Solution: We have
1
z /2
=
M361 (56650) Midterm 1 Solutions
1. (10 points) Suppose that cos z = 0. Prove that z is real.
Solution: Suppose z C and cos z = 0. Then eiz + eiz = 0, so e2iz = 1.
Write z = x + iy with x, y R. We have
1 = e2iz = e2ix2y = e2y cos 2x + ie2y sin 2x.
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Result of who makes
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Collectors who buy
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Japanese Culture
Some art is not mea
Math 362K Probability
Fall 2007
Instructor: Geir Helleloid
Practice Midterm 1: Solutions
1. In a group of 3120 entering freshmen, 3002 have 2 years of algebra, 2962 have two
years of English, and 2340 have two years of languages. If 2902 have both algebra
M361 (56650) Final exam
Name (print):
EID:
This exam is 7 pages long. There are 9 questions. Answer all questions in the space
provided. Phones, calculators, notes, and other aids are forbidden.
Question:
1
2
3
4
5
6
7
8
9
Total
Points:
10
10
10
10
10
10
M361 (56650) Final exam Solutions
1. (10 points) Let f (z ) = ez . Show that f is not holomorphic.
Solution: We have
f (z ) = ez = ex cos y iex sin y.
Then
fx = ex cos y iex sin y
fy = ex sin y iex cos y,
so fx = fy /i. Since f doesnt satisfy the Cauchy-R
Statistics 100A
Homework 4 Solutions
Ryan Rosario
Chapter 4
39. A ball is drawn from an urn containing 3 white and 3 black balls. After the ball is drawn, it
is then replaced and another ball is drawn. This goes on indenitely. What is the probability
that
Math361 HW # 1
Jonathan Campbell
Due: 1/28/2016
Problem 1 Heres some practice computing with complex numbers.
1. Express, with no complex denominator,
2
5 12i
5 + 5i
2 + 3i
1i
1+i
2. Write ( 3 + i)78 as a complex number in the form x + iy.
3. Convert 10ei
Math361 HW # 2
Jonathan Campbell
Due: 2/04/2016
Problem 1 Show that a triangle with vertices a, b, c is equilateral if
a + b + c 2 = 0
where is the 3rd root of unit e2i/3 .
Problem 2 Recall that the centroid of a triangle is the intersection of the lines
Math361 HW # 4
Jonathan Campbell
Due: 2/18/2016
Problem 1 Let : [a, b] be a curve. Show that (t) = (b t) + a) is that same
curve but parameterized backwards. For f : C a complex function,
show
Z
Z
f (z) dz = f (z) dz
Problem 2 Show that if |a| < r < |b| t
Math 361 HW 5
Due: March 3, 2016
Problem 1 Compute
Z
sin 5z
dz
z + /2
where is cfw_|z| = 5.
Problem 2 Compute
eiz
dz
z3
Z
where = cfw_|z| = 2.
Problem 3 Compute
1
2i
Z
z 2 dz
z2 + 4
where is a square with vertices at 2 and 2 + 4i.
Problem 4 Show that
Z
2
Math361 HW # 3
Jonathan Campbell
Due: 2/11/2016
Problem 1 Show that
1
f (z) =
2
1
z+
z
is a conformal map from the upper half disc (D H) to the upper half
plane.
Problem 2 Use the previous exercise to compute where sin z takes the strip
S = cfw_z : Im(z)
Math361 HW # 6
Jonathan Campbell
Due: 3/10/2016
Problem 1 Show
2
Z
2
d
=
a + b sin
a2 b2
0
Problem 2 Show that
Z
Problem 3 Compute
Z
0
Problem 4 Evaluate
Z
0
Problem 5 Prove that
2
Z
0
Problem 6 Evaluate
1
2i
d
= 2
2
1 + sin
Z
C
(x2
dx
+ 1)(x2 + 4)2
cos
Math361 HW # 7
Jonathan Campbell
Due: 3/31/2016
Problem 1 Find an expression for the sum of the series
X
n2 z n
|z| < 1.
n=1
Hint: it is the derivative (or perhaps second derivative of a certain series).
Justify convergence.
Problem 2 Use the Taylor serie
Math361 HW # 8
Jonathan Campbell
Due: 4/14/2016
Problem 1 In class, I used the fact that
Lcfw_ty 0 (t)(s) = Y (s) sY 0 (s)
Prove this.
Problem 2 Show that
Lcfw_ty(t) =
d
Y (s)
ds
Problem 3 Let Hn (x) be the Hermite polynomials derived in class, i.e.
n
2
M361 (56650) Problem Set 2 Solutions
1. Suppose that C is open and connected and f : C is holomorphic.
(a) Show that if Re f is constant, then f is constant.
Solution: Write f = u + iv . We have assumed that u is constant, so ux = uy =
0. Since f is holom
M361 (56650) Problem Set 3 Solutions
1. Let n 0 be an integer. The function
f (z ) = (1 + x2 )n1
has poles at i.
(a) Compute ordi f and ordi f .
Solution: Since
(1 + x2 ) = (x + i)(x i),
we have ordi f = ordi f = n + 1.
(b) Compute resf i and resf i.
Solu
M361 (56650) Problem Set 5 Solutions
1. For each of the following dierential equations, nd the solution in terms of z (0) and
sketch the orbits of the corresponding ow in C:
(a) Let > 0. Solve the logisitic equation z = z (1 z ).
Solution: There are many
M361 (56650) Problem Set 4 Solutions
All exercises are taken from Plya-Latta, 1974.
o
1. Suppose a is real and a2 < 1. Compute the integral
cos 2
d
1 2a cos + a2
0
Solution: We write
I=
0
Observe that
cos 2
d.
1 2a cos + a2
2I =
cos 2
d
1 2a cos + a2
2
si
M361 (56650) Problem Set 3 Solutions
1. Let n 0 be an integer. The function
f (z ) = (1 + x2 )n1
has poles at i.
(a) Compute ordi f and ordi f .
Solution: Since
(1 + x2 ) = (x + i)(x i),
we have ordi f = ordi f = n + 1.
(b) Compute resf i and resf i.
Solu
M361 (56650) Problem Set 2 Solutions
1. Suppose that C is open and connected and f : C is holomorphic.
(a) Show that if Re f is constant, then f is constant.
Solution: Write f = u + iv . We have assumed that u is constant, so ux = uy =
0. Since f is holom
M361 (56650) Final exam Solutions
1. (10 points) Let f (z ) = ez . Show that f is not holomorphic.
Solution: We have
f (z ) = ez = ex cos y iex sin y.
Then
fx = ex cos y iex sin y
fy = ex sin y iex cos y,
so fx = fy /i. Since f doesnt satisfy the Cauchy-R
M361 (56650) Problem Set 5 Solutions
1. For each of the following dierential equations, nd the solution in terms of z (0) and
sketch the orbits of the corresponding ow in C:
(a) Let > 0. Solve the logisitic equation z = z (1 z ).
Solution: There are many
M361 (56650) Problem Set 4 Solutions
All exercises are taken from Plya-Latta, 1974.
o
1. Suppose a is real and a2 < 1. Compute the integral
cos 2
d
1 2a cos + a2
0
Solution: We write
I=
0
Observe that
cos 2
d.
1 2a cos + a2
2I =
cos 2
d
1 2a cos + a2
2
si
M361 (56650) Problem Set 3 Solutions
1. Let n 0 be an integer. The function
f (z ) = (1 + x2 )n1
has poles at i.
(a) Compute ordi f and ordi f .
Solution: Since
(1 + x2 ) = (x + i)(x i),
we have ordi f = ordi f = n + 1.
(b) Compute resf i and resf i.
Solu
M361 (56650) Problem Set 2 Solutions
1. Suppose that C is open and connected and f : C is holomorphic.
(a) Show that if Re f is constant, then f is constant.
Solution: Write f = u + iv . We have assumed that u is constant, so ux = uy =
0. Since f is holom