Math 372: Fall 2015: Solutions to Homework
Math 372: Fall 2015: Solutions to Homework
Steven Miller
December 7, 2015
Abstract
Below are detailed solutions to the homework problems from Math 372 Complex Analysis (Williams College,
Fall 2015, Professor Steven J. Miller, sjm1@williams.edu). The course homepage is
and the textbook is Complex Analysis by Stein and Shakarchi (ISBN13: 978-0-691-11385-2). Note to students:
it¡¯s nice to include the statement of the problems, but I leave that up to you. I am only skimming the solutions.
I will occasionally add some comments or mention alternate solutions. If you find an error in these notes,
let me know for extra credit.
Contents
1 Math 372: Homework #1: Yuzhong (Jeff) Meng and Liyang Zhang (2010)
1.1 Problems for HW#1: Due September 21, 2015 . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.2 Solutions for HW#1: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3
3
3
2 Math 372: Homework #2: Solutions by Nick Arnosti and Thomas Crawford (2010)
8
3 Math 372: Homework #3: Carlos Dominguez, Carson Eisenach, David Gold
4 Math 372: Homework #4: Due Friday, October 12, 2015: Pham, Jensen, Kolog?lu
4.1 Chapter 3, Exercise 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.2 Chapter 3, Exercise 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.3 Chapter 3, Exercise 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.4 Chapter 3 Exercise 15d . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.5 Chapter 3 Exercise 17a . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.6 Additional Problem 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
12
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
16
16
17
19
22
22
22
5 Math 372: Homework #5: Due Monday October 26: Pegado, Vu
24
6 Math 372: Homework #6: Kung, Lin, Waters
34
7 Math 372: Homework #7: Due Monday, November 9: Thompson, Schrock, Tosteson
42
7.1 Problems. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46
1
8 Math 372: Homework #8: Thompson, Schrock, Tosteson
47
8.1 Problems. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47
8.2 Solutions. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47
9 Math 372: Homework #9: Miller, Xiong, Webster, Wilcox
2
49
1 Math 372: Homework #1: Yuzhong (Jeff) Meng and Liyang Zhang (2010)
1.1 Problems for HW#1: Due September 21, 2015
Due September 21: Chapter 1: Page 24: #1abcd, #3, #13.
Problem: Chapter 1: #1: Describe geometrically the sets of points z in the complex plane defined by the following relations: (a) |z ? z1 | = |z ? z2 | where z1 , z2 ¡Ê C; (b) 1/z = z; (c) Re(z) = 3; (d) Re(z) > c (resp., ¡Ý c)
where c ¡Ê R.
Problem: Chapter 1: #3: With ¦Ø = sei? , where s ¡Ý 0 and ? ¡Ê R, solve the equation z n = ¦Ø in C where n is a
natural number. How many solutions are there?
Problem: Chapter 1: #13: Suppose that f is holomorphic in an open set ?. Prove that in any one of the following cases f must be constant:
(a) Re(f ) is constant;
(b) Im(f ) is constant;
(c) |f | is constant.
1.2 Solutions for HW#1:
Due September 21, 2015: Chapter 1: Page 24: #1abcd, #3, #13.
Problem: Chapter 1: #1: Describe geometrically the sets of points z in the complex plane defined by the following relations: (a) |z ? z1 | = |z ? z2 | where z1 , z2 ¡Ê C; (b) 1/z = z; (c) Re(z) = 3; (d) Re(z) > c (resp., ¡Ý c)
where c ¡Ê R.
Solution: (a) When z1 6= z2 , this is the line that perpendicularly bisects the line segment from z1 to z2 . When
z1 = z2 , this is the entire complex plane.
(b)
1
z
z
=
= 2.
(1.1)
z
zz
|z|
So
1
z
= z ? 2 = z ? |z| = 1.
z
|z|
(1.2)
This is the unit circle in C.
(c) This is the vertical line x = 3.
(d) This is the open half-plane to the right of the vertical line x = c (or the closed half-plane if it is ¡Ý).
Problem: Chapter 1: #3: With ¦Ø = sei? , where s ¡Ý 0 and ? ¡Ê R, solve the equation z n = ¦Ø in C where n is a
natural number. How many solutions are there?
3
Solution: Notice that
¦Ø = sei? = sei(?+2¦Ðm) , m ¡Ê Z.
(1.3)
It¡¯s worth spending a moment or two thinking what is the best choice for our generic integer. Clearly n is a bad
choice as it is already used in the problem; as we often use t for the imaginary part, that is out too. The most natural
is to use m (though k would be another fine choice); at all costs do not use i!
Based on this relationship, we have
z n = sei(?+2¦Ðm) .
(1.4)
So,
z = s1/n e
i(?+2¦Ðm)
n
.
(1.5)
Thus, we will have n unique solutions since each choice of m ¡Ê {0, 1, . . . , n ? 1} yields a different solution so
long as s 6= 0. Note that m = n yields the same solution as m = 0; in general, if two choices of m differ by n then
they yield the same solution, and thus it suffices to look at the n specified values of m. If s = 0, then we have only
1 solution.
Problem: Chapter 1: #13: Suppose that f is holomorphic in an open set ?. Prove that in any one of the following cases f must be constant:
(a) Re(f ) is constant;
(b) Im(f ) is constant;
(c) |f | is constant.
Solution: Let f (z) = f (x, y) = u(x, y) + iv(x, y), where z = x + iy.
(a) Since Re(f ) = constant,
?u
?u
= 0,
= 0.
?x
?y
(1.6)
By the Cauchy-Riemann equations,
?v
?u
=?
= 0.
?x
?y
Thus, in ?,
f ¡ä (z) =
?f
?u
?v
=
+i
= 0 + 0 = 0.
?x
?x
?x
(1.7)
(1.8)
Thus f (z) is constant.
(b) Since Im(f ) = constant,
?v
?v
= 0,
= 0.
?x
?y
(1.9)
?u
?v
=
= 0.
?x
?y
(1.10)
By the Cauchy-Riemann equations,
Thus in ?,
f ¡ä (z) =
?f
?u
?v
=
+i
= 0 + 0 = 0.
?x
?x
?x
4
(1.11)
Thus f is constant.
¡Ì (c) We first give a mostly correct argument; the reader should pay attention to find the difficulty. Since |f | =
u2 + v 2 is constant,
(
2
2
?v
0 = ?(u?x+v ) = 2u ?u
?x + 2v ?x .
(1.12)
2
2
?v
0 = ?(u?y+v ) = 2u ?u
?y + 2v ?y .
Plug in the Cauchy-Riemann equations and we get
u
?v
?v
+v
= 0.
?y
?x
?u
(1.13)
?v
?v
+v
= 0.
?x
?y
(1.14) ?
v ?v
?v
=
.
?x
u ?y
(1.14)
(1.15)
Plug (1.15) into (1.13) and we get
u2 + v 2 ?v
= 0.
u
?y
(1.16)
So u2 + v 2 = 0 or ?v
?y = 0.
2
2
If u + v = 0, then, since u, v are real, u = v = 0, and thus f = 0 which is constant.
Thus we may assume u2 + v 2 equals a non-zero constant, and we may divide by it. We multiply both sides by
?v
?u
u and find ?v
?y = 0, then by (1.15), ?x = 0, and by Cauchy-Riemann, ?x = 0.
f¡ä =
?f
?u
?v
=
+i
= 0.
?x
?x
?x
(1.17)
Thus f is constant.
Why is the above only mostly a proof? The problem is we have a division by u, and need to make sure
everything is well-defined. Specifically, we need to know that u is never zero. We do have f ¡ä = 0 except at points
where u = 0, but we would need to investigate that a bit more.
Let¡¯s return to
(
2
2
?v
0 = ?(u?x+v ) = 2u ?u
?x + 2v ?x .
(1.18)
2
2
?v
0 = ?(u?y+v ) = 2u ?u
?y + 2v ?y .
Plug in the Cauchy-Riemann equations and we get
?v
?v
+v
?y
?x
?v
?v
?u
+v
?x
?y
u
5
=
0
=
0.
(1.19)
................
................
In order to avoid copyright disputes, this page is only a partial summary.
To fulfill the demand for quickly locating and searching documents.
It is intelligent file search solution for home and business.
Related download
- department of civil engineering cve 372 hydromechanics
- the bivariate normal distribution
- comfort grundfos
- exercise problems open channel flow
- 1 organizationofcitygovernment 372 iowa
- edta titration calculations community college of rhode
- 1 organizationofcitygovernment 372
- machine screws thread dimensions aspen fasteners
- chapter 6 linear transformation university of kansas
- metric thread extended thread size range
Related searches
- answers to homework questions free
- answers to homework assignments
- answer to homework questions
- solutions to biodiversity loss
- solutions to lowering college tuition
- complex solutions to quadratic equations
- solutions to quadratic equations calculator
- solutions to quadratic equations pdf
- solutions to inequalities calculator
- integer solutions to inequalities calculator
- solutions to the inequality calculator
- solutions to water pollution