1549312215-Complex_Analysis_for_Mathematics_and_Engineering_5th_edition__Mathews

594 ANSWERS

(show t he details for this). But this last quantity equals J3 -v'6 (explain),

which is less than 1 (again, explain).

1 3. Since If' (zo)I < 1, we can choose p such that If' (zo)I < p < 1. Using the

same technique as Theorem 4.10, show that if z ED; (zo), then If (z) -zol <

p lz -zol· That is, lz1 -zol < p lz -zol, where z1 = f (z). An easy induction

argument now gives that for all k, lzk -zol < p" lz - zol, where zk is the kth

iterate of z. Since p < 1, t his implies lim zk = zo. Show t he details.

k-oo

Section 4.3. Geometric Series and Convergence Theorems: page 146

00 (l+i)n

1. By Theorem 4.12, L: z;;-= (~) = 1 + i (show the details), since

n=-0 1- 2

1 1¥ 1 < 1 (show this also).

3. The series converges by the ratio test. Show the details.

5a. Converges in D v; (0).

T

5c. Converges in D; (i).

7. ISnl = I 1 !., - (~.1 ~ I (~.1- 11 .:. 1 = lz" I J 1 !.. 1-I 1 !., J. Now use the fact
   that lzl > 1 to get the desired conclusion.


8. Mimic the argument most calculus texts give for real series, but replace lxl
   with lzl.

11. If f (z) = E z (zn), then f (z^2 ) = E (z^2 )(^2 "l = E z<^2 ·^2 "> = E z(^2 "+') =

n=O n=O n=O n= O

00

L: z<^2 ">. The conclusion follows from this. Explain in det ail, especially t he

n=;l

second equality for f (z^2 ).

Section 4.4. Power Series Functions: page 153

1. T he series for j (z) converges absolutely if nlim -oo I c~+rt ' I lz - a l < 1.

If Jim I c~.,_, I = 0, the series converges for all z. If Um I c~+^1 I = oo, the

n - oo n n-oo "

series converges only when z = a. If J~n;, I~' I is finite but not zero, then

the series converges if lz - a l < lim 11 ;:.,.±.L I = P·

.i-oo en

3a. oo.