ANSWERS 60500
15b. f (z) = l + l L 2-n (zn + z - n).
n = l
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17. Since f (z) = L anzn is valid for lzl = 1 (explain), letting z = ei^8
n=-oo
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gives f ( ei^8 ) = I: anein^8 immediately. By Laurent's theorem, an =
n=-ooz~i fct fJ~~ dz for all integers n (explain). Parametrizing Ci (O) with
z ( 8) = ei<P for 0 5 </> 5 211" gives the desired result. Show the det ails.
00- Since L C- n (z - a:)- n converges for lz - a l > r, the ratio test guarantees
n=l
that the series converges absolutely for { z : lz - al :2: s}, where s > r (show
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t he details). Thus, if lz - a:I = s, the series L lc- nl s-n converges. Since
n = l
lc-n (z - oT"I 5 lc-nl s - n for all lzl :2: s, the Weierstrass M -test gives us
our conclus ion. Explain.Section 7.4. Singularities, Zeros, and Poles : page 283
la. Zeros of order 4 at ± i.l e. Simple zeros at -1 ± i.
le. Simple zeros at ± i and ±3i.lg. Simple zeros at '1±•, -q±;, and ±i.
li. Zeros of order 2 at 1±;"3 and -1.Simple zeros at^1 ji and -~i, and a zero of order 4 at the origin.
2a. Poles of order 3 at ±i, and a pole of order 4 at 1.2c. Simple poles at q±;, -q±•, and ± i.
2e. Simple poles at ±./3i and ~-2g. Simple poles at z = n7r for n = ±1, ±2,...2i. Simple poles at z = n7r for n = ±1, ±2, ... , and a pole of order 3 at the
origin.2k. Simple poles at z = 2n11" for n = 0, ±1, ±2,...
3a. Removable singularity at the origin.3c. Essential singularity at the origin.