Gamelin, Problem 11, VI.2
May 18, 2014
This note contains a solution of Problem 11, Section VI.2 of Gamelin, [1].
X
Problem. Suppose f (z) =
ak z k is analytic for |z| < R, and suppose that f extends to be meromorphic
for |z| < R + , with only one pole z0 on the circle |z| = R. Show that ak /ak+1 → z0 as k → ∞.
Solution. Let
f (z) −
m
X
j=1
X
bj
=
ck z k
(z − z0 )j
be analytic in |z| < R + . Then using the binomial theorem, or otherwise
ak = ck +
m
X
(−1)j
j=1
m
X
ak+1 = ck+1 +
j=1
z0j
bj
(−1)j
z0j
(k + j − 1) . . . (k + 1)
z0k+j−1
bj
(k + j) . . . (k + 2)
z0k+j
.
Dividing and multiplying top and bottom by z0k+1 we get
P
1−2j
j
ck z0k + m
ak
j=1 (−1) bj (k + j − 1) . . . (k + 1)z0
= z0
P
1−2j
j
ak+1
ck+1 z0k+1 + m
j=1 (−1) bj (k + j) . . . (k + 2)z0
= z0
Since
X
ck z0k + (−1)m bm z01−2m k m−1 + . . .
.
ck+1 z0k+1 + (−1)m bm z01−2m k m−1 + . . .
ck z0k converges, ck z0k → 0 as k → ∞. Now let k → ∞ and see that
ak
→ z0 .
ak+1
References
[1] Theodore Gamelin, Complex Analysis, Springer-Verlag, 2001.
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