590 ANSWERS
15. T he intersection of D~ (~) = { (u, v): (u- !)
2+ v^2 < ~} and Dt (-~) =
{Cu, v): u^2 +(v+ ~)
2
< !}·- The map w = -1 + ~ (with inverse z = w!i) has lz - 11< 1 <===>
I W~I - 11 < 1 <===> I u+i~+l - 11 < 1 <===> I z~~i~;-:~i - 11 < 1. Amazingly,
this simplifies to 4u [ ( u + 1)^2 + v^2 ] > 0, which occurs iff u = Re (w) > 0.
Show the details!19. Let e > 0 be given, Choose R = ~ + l. Assume lzl > R = ~ + 1. Then
lz - 11 ~ lzl- 1 > (~+1)- 1 = : .Therefore, 1, : 11 < e, so 1 ~ -11 =\ • .:. 1 \ < e. To see how to get R, start wit h I~ -1j < e, and work back-
wards.- Broadly speaking, ±oo are designations for limits in calculus indicating quan-
tities that get arbitrarily positive or negative. There is no such measure in
complex analysis. Further, t he point oo can be given a meaningful definition
on the Riemann sphere. There is no such analogy for ±oo. Elaborate and
give some other comparisons. The sector p > 0, ~ < <P < ~-
Section 3.1. Differentiable and Analytic Functions: page 98
la. f'(z) = 15z^2 -8.z+7.
le. h' (z) = (z: 2 >2 for z f -2.
- Parts (a), (b), (e), (f) are entire, and (c) is entire provided that g (z) f 0 for
all z.
5. The result is clearly true when n = 1. Assume for some n > 1 tha t P' (z) =
n+I na1 +2a2z+ · · · + na,,zn-^1. consider Q (z) = L: akzk = L: akzk+ an+lzn+l.
k=O k=O
Since the derivative of t he sum of two terms is t he sum of t he derivatives, we
have Q' (z) = f, ( f: akzk) + f, (an+1z"+^1 ). The induction assumption
k = O
now gives t he required result.7a. -4i.7c. 3.7e. -16.- f. z - n = -J. (.~). Apply t he quotient rule f. (.~) = {zn};f.~~~l~ -k<•n) and
simplify.