1549312215-Complex_Analysis_for_Mathematics_and_Engineering_5th_edition__Mathews

604 ANSWERS

h (•)(<>) 13! (•)(<>)

13. To verify Identity (7- 15 ), let h(z) = f3f (z). Clearly, ----;ir-= n! =

00

f3an· By Taylor's theorem, h (z) = /3! (z) = L /Jan (z -at.

n=O

15. Use the fact that f'(z) = [z- (-1 +i) + (-1 +i))-

1

and expand f'(z) in

powers of [z - (- 1 + i)]. Then apply Corollary 7.2.

17a. By definition, J(- z) = -f(z), so using t he chain rule, we see that f'(z) =
f,J(z) = -j,J(-z) = -f'(- .z)(-1) = f'(- z). But this means that f'
is an even function.

17c. If f is even, then by part bf' is odd, so f'(O) = -f'(-0) = -f'(O).

Of course, this implies f' (0) = 0. Similarly, from part a f" is even, so

fm (0) = O. An induction argument gives f (zn-l) (0) = 0 for all positive

integers n. Show the details.

19a. It is easy to show that f (n) (0) = n! for all positive integers n. Do so via

mathematical induction.

19b. The point z =! is a removable singularity, since f may b e redefined at!

to be analytic. State what f should equal at that point.

Section 7 .3. Laurent Series Representations: page 274

00 00

1. z3:,a = L zn-^3 for 0 < lzl < 1, •• :,, = - L )i:• for lz l > 1.

n = O n=l

00 ( l)"-z2rt+l z2n-3

3. L - (2n+1)! for lzl > 0.

n=O

00 ( l)"

5. L (2n+~)lz2n+I valid for lzl > o.
   n=O
   oo 2z""-7


6. L ( 4 n-z)! valid for lzl > O.
   n = l


7. z -1 (4 - z) -2-_ I oo (n+~;•
   16 , + n=O L 4 • for lzl < 4,

z-^1 (4 - z) - 2 = L^00 n(z •^4 >;;• for lzl > 4.

n=I

( )

00 n n

11. Log ~=~ = L b n~~ valid for lzl > b. Explain.

n = l

(^13) .CSCZ_ -;^1 + z s +J60 7z3 + ·· ·.
15a. This identity is obtained by straightforward subsitution, and partial fraction
decomposition.