Power Series Coefficients and Poles

Question

Say we have a function $f(z)$ which is holomorphic at $ z = 0 $ but meromorphic at $ z = a $ with a simple pole at this point. The function is holomorphic elsewhere in the circle $ |z|<R$ where $R > a$.

How would you go about showing that the coefficients $c_n$ of the Taylor Series about $z=0$ grow as $ c_n \sim \frac{1}{a^n} $? And what would change in the analysis if the pole was made to be of order $d>1$?

I've tried using the Cauchy integral formula for $z=0$ but to no avail. Any hints or tips would be much appreciated.

Answer 1

Let $f(z)= \sum_{n=0}^{\infty}c_nz^n$ the Taylor series about $z=0.$ The radius of convergence is exactly $|a|$. Hence

$$ \lim \sup |c_n|^{1/n}= \frac{1}{|a|}.$$

Can you proceed ?

Answer 2

For simplicity let's assume that the function $f$ is meromorphic with a single pole of order $d+1$ at $a$. Let $\alpha_n$ denote its Taylor coefficients at the origin. We define $$ g(z) = f(z) - \frac{c}{(1-x/a)^{d+1}}, $$ with $c$ the same as the pole's constant. Let $\beta_n$ denote its Taylor coefficients at the origin.

Now take Cauchy's integral $$g(z)=\oint\frac{g(\zeta)}{\zeta-z}d\zeta$$ and you can prove that $g$ is holomorphic.

Cauchy's integral also gives bounds for the Taylor coefficients of $g$. By differentiating Cauchy's integral we get $$\beta_n=\frac{g^{(n)}(0)}{n!} =\oint\frac{g(\zeta)}{\zeta^{n+1}}d\zeta.$$ Now choose $r>|a|$ and let $M>0$ be such that if $|z|\le r$ then $|g(z)|\le M/2\pi$. Then we take the above relation, we set the contour of integration to be the circle $|z|=r$ and we take the absolute value: $$|\beta_n|\le \oint\frac{|g(\zeta)|}{|\zeta|^{n+1}}|d\zeta|\le Mr^{-n}.$$

For the pole we start with $$\frac{1}{1-z} = \sum_{n\ge 0} z^n.$$ We differentiate this relation $d$ times and we get $$\frac{1}{(1-z)^{d+1}} = \sum_{n\ge 0} \binom{n+d}{n} z^n.$$ Then $$\frac{1}{(1-z/a)^{d+1}} = \sum_{n\ge 0} \binom{n+d}{n} a^{-n} z^n.$$

Now we put everything together. We know that $f(z)=g(z) + \frac{c}{(1-x/a)^{d+1}}$, so $$ \alpha_n = \beta_n + \binom{n+d}{n} a^{-n}. $$ This means that $$ |\alpha_n| \le Mr^{-n} + \binom{n+d}{n} |a|^{-n}. $$ Since $|a|<r$, there exists $N>0$ such that for all $n\ge0$, $$N \binom{n+d}{n} |a|^{-n} \ge Mr^{-n}.$$ From this we finally get that $$ |\alpha_n| \le (N+1)\binom{n+d}{n} |a|^{-n}. $$

You can easily adapt the above for functions meromorphic in a disk with multiple poles.