1549312215-Complex_Analysis_for_Mathematics_and_Engineering_5th_edition__Mathews

606 ANSWERS

3e. Removable singularity at the origin, and a simple pole at -1.

3g. Removable singularity at the origin.

5. By Theorem 7.11, f (z) = (z - zo)k h (z), where h is analytic at zo and
   h (zo) 'f O. We compute

J' (z) = k (z -zo)k- l h (z) + (z - zo)k h' (z)

= (z -zo)k- l [kh (z) + (z - zo) h' (z)]

= (z - zo)k-1 g (z)'

where g (z) = kh(z) + (z -zo) h' (z). Explain why g (zo) ¥ 0, why g is

analytic at z 0 , and why Theorem 7.11 now gives the conclusion.

7. If it so happens that m = n , and the coefficients in the Laurent expansions

for f and g about zo are negatives of each other, then f + g will have a

Taylor series representation at zo, making zo a removable singularity (show

the details for this). If m ¥ n, then it is easy to show that f + g still has a

pole. State why, and what the order of the pole is.

9. Appeal to Theorem 7.12 and mimic the argument given in the solution to
   Problem 5.

lla. Simple poles at z = n'" for n = ±1, ±2, ... , and a nonisolated singularity at

the origin.

Section 7.5. Applications of Taylor and Laurent Series: page 289

la. No. Otherwise 0 = Jim f ( 2 ~) = f ( Jim 2 ~) = f (0). On the other hand,

n -oo n -oo

1 = n-oo Jim f ( 2 n1 1 ) = f ( n-+oo Jim (^2) fl 1 1 ) = f (0). Justify and explain.
lb. Yes. There is a simple function with this property. Find it.
le. No. Use Corollary 7.10 to show that for all .z in some disk Dr (0) we have
f ( z) = z^3 , and f ( z) = -z^3 , and explain why this is impossible.

3a. Let Zn = "~. Explain.

3b. No, the function f is not analytic at zero (explain why), which is required

by the corollary.

5. For x '/ 0, Jim lxsin~I ~Jim lxl = O. This implies Jim f (x) = 0 = f (0).

x-+O x-0 x-o

For the complex case, show that there is an essential singularity at O and

use Theorem 7.17.