MA 513 Complex Variables
First Test Solutions
February 2009
1. (20 points) (a) Find all solutions of the equation z 3 + 1 i = 0. Answer: 2 1/6 1/6 1/6 z = 22 (1+i); z = 21/6 ei( 4 3 ) : z = 22 (1 3+i( 31); z = 22 (1+ 3i( 3+1) 2 2 (b) Find all the numbers
MA 513 Homework 8 Solutions
p.255: #5; p267: #1,5,6; p. 275: #4,7; p.286: #2.
1
p. 255: #5: Within the contour ON the function f = Z2 Sinz had singularities z = 0,2 =
imr, n 2 17 2, ...N. At 2: = 0, sinz = z 23/6 + ..., so that f(z) has a pole of
MA 513
Homework 8
Solutions
p.255: #5; p.267: #1,5,6; p. 275: #4,7; p.286: #2. p. 255: #5: Within the contour CN the function f (z ) = n, n = 1, 2, .N. At z = 0, sin z = z z 3 /6 1 z 2 sin z had singularities z = 0, z =
+ ., so that f (z ) has a pole of o
MA 513
Homework 7
Solutions
p.239: #1, 2; p.243: #1, 2; p.248: #1,3. p. 239: #1: (a) 1 1 = .Res = 1. 2 z+z z (z + 1)
(b) z cos(1/z ) = z (1 (1/z )2 /2 + .) = 1/2z + series with other powers. Res = 1/2. (c) z sin z 1 = (z (z z 3 /6 + .), so there is no 1/z
MA 513
Homework 6
Solutions
p.188: #4; p.197: #8, 13; p. 205: #2, 3, 4; p. 220: #4. p. 188: #4: For 0 < r < 1,
(rei )k =
k =0 k =0
r k (cos k+i sin k) =
1 1 1 r cos ir sin . = = i 1 re 1 r cos ir sin (1 r cos )2 + r 2 sin2
Now equate real and imaginary
MA 513 p. 170: #5, 7, 10; p. 178: # 1, 2, 9.
Homework 5
Solutions
p. 170: #5:
C
f (z ) dz = 2if (z0 ) = z z0
C
f (z ) dz . (z z0 )2
#7: so
C
eaz dz = 2ieaz z
= 2i, by the Cauchy Integral Formula. Now z = cos + i sin ,
z =0
C
eaz dz = z =
exp(a(cos + i s
MA 513
Homework 4
Solutions
p.136: #11; p.140: #1,3,4; p.149: #2(a); p.160: #1(a)(e)(f); p.170: #1(a), 2,3. p.136: # 11: Let t = ( ). Then dt = ( )d. Therefore, with Z ( ) = z ( ), we have Z ( )d = z ( ) ( )d = z (t)dt. p.140: #1: Since |z 2 1| |z |2 1| =