ANSWERS 603there exists an integer N. such that n 2': N. implies lln (z) - I (z)I < ~ for
all z E T. Likewise, there exists an integer M. such that n 2': M. implies
IYn (z) - g (z)I < ~ for all z ET. If we set L£ =Max {N., M.}, then for n 2':
L,, l(fn (z) + 9n (z)) - (f (z) + 9 (z))I $ lfn (z) - I (z)I + l9n (z) - g (z)I <
~ + ~ = e for all z E T.7b. For all n, let In (x) = x, and 9n (x) = ~. for all x E T, where Tare the
real numbers. Then In (x) converges uniformly to x, and 9n (x) converges
uniformly to 0 (verify). However, even though In (x) Yn (x) converges to 0
(explain), the convergence is not uniform (verify). Can you come up with a
different example?9a. For z E A, ln-•I = lexp [-(x + iy) lnnJI = lexp (-iy ln n)l lexp (-xlnn)I =
n-"'. Since z E A, we know Re(z) = x 2': 2, so n-x $ tz· Thus, with
Mn = ,&, we see that ((z) converges uniformly on A by the Weierstrass
M-test.Section 7.2. Taylor Series Representations: page 264
. oo z2n.+ t
l a. smhz = I: ( 2 n+l)! for all z.
n=O
00 (-""-'
l e. Log(l + z) = I: ~z" for all z E D 1 (0).
n=l3a. !:::~ = 1 _'(; : 1 ) = (z -1) [ l -(!-i)]. Expand the expression in brackets by
replacingz withz- 1 in thegeometricseries(valid, therefore, for lz-11<1),
then multiply by t he ( z - 1) term.- I (z) = 1 !.. = ,:_. [ i-s J. Expand the expression in brackets by replacing
z with ~=~ in the geometric series (valid, therefore, for I~:::! I < 1, or lz -ii <
v'2). Explain.7a. By Taylor's theorem, '1~/^0 > = (3 + (-1)")". Therefore, 1 <~/0l = 8, so
I <^3 > (O) = 48.
00 00
9a. Observe that l +zl (z)+z^2 I (z) = 1 + I: c,..zn+l + I: c,,zn+^2. Reindex and
n=O n : O
00 00 00
write this as 1+ L Cn-1zn+ I: Cn-2z" = l+z+ I: (Cn-1 + Cn-2) z". Now
n=l n:2 n : 2use the relation Cn = Cn-i +Cn- 2 for n 2': 2 to conclude l+zl (z)+z^2 I (z) =
I (z). Solve for f (z).
- The point z is on the circle Gp (a) with center a, so z f a. Also, zo is in
the interior of this circle, so again z =f zo.