Gamelin, Problem 11, VI.2

May 18, 2014

This note contains a solution of Problem 11, Section VI.2 of Gamelin, [1]. Problem. Suppose f(z) = ∑ a_{k}z^{k} is analytic for |z| < R, and suppose that f extends to be meromorphic for |z| < R + ϵ, with only one pole z_{0} on the circle |z| = R. Show that a_{k}/a_{k+1} → z_{0} as k → ∞. Solution. Let

f(z) − ∑_{j=1}^{m} \frac{b_{j}}{(z − z_{0})^{j}} = ∑ c_{k}z^{k}

be analytic in |z| < R + ϵ. Then using the binomial theorem, or otherwise

a_{k} = c_{k} + ∑_{j=1}^{m} \frac{(−1)^{j}}{z_{0}^{j}} b_{j} \frac{(k + j − 1) … (k + 1)}{z_{0}^{k+j−1}}
a_{k+1} = c_{k+1} + ∑_{j=1}^{m} \frac{(−1)^{j}}{z_{0}^{j}} b_{j} \frac{(k + j) … (k + 2)}{z_{0}^{k+j}}.

Dividing and multiplying top and bottom by z_{0}^{k+1} we get

\frac{a_{k}}{a_{k+1}} = z_{0} \frac{c_{k}z_{0}^{k} + ∑_{j=1}^{m}(−1)^{j}b_{j}(k + j − 1) … (k + 1)z_{0}^{1−2j}}{c_{k+1}z_{0}^{k+1} + ∑_{j=1}^{m}(−1)^{j}b_{j}(k + j) … (k + 2)z_{0}^{1−2j}}

= z_{0} \frac{c_{k}z_{0}^{k} + (−1)^{m}b_{m}z_{0}^{1−2m}k^{m−1} + …}{c_{k+1}z_{0}^{k+1} + (−1)^{m}b_{m}z_{0}^{1−2m}k^{m−1} + …}.

Since ∑ c_{k}z_{0}^{k} converges, c_{k}z_{0}^{k} → 0 as k → ∞. Now let k → ∞ and see that

\frac{a_{k}}{a_{k+1}} → z_{0}.

References

[1] Theodore Gamelin, Complex Analysis, Springer-Verlag, 2001.