I am currently reviewing some complex analysis, and have come across this question which I absolutely have no idea on how to attempt:
Suppose the radius of convergence of the power series $f(z) = \sum_{n = 0}^{\infty}a_nz^n$ is $1$, and $f$ has only finitely many singularities $z_1,\ldots,z_m$ on the unit circle $|z| = 1$ which are all simple poles. Show $\{a_n\}$ is bounded.
Obviously it suffices to show $\sum_{n = 0}^{\infty}a_n < \infty$. Since there are only finitely many singularities, we can find a $w_1$ with $|w_1| = 1$ such that $\sum_{n = 0}^{\infty}a_nw_1^n < \infty$…? But I have no intuition on how to proceed, since most of the theory I've read is for points interior to the circle of convergence.
Any suggestions appreciated!
Best Answer
The thought process should probably begin with: let's look at some function with a simple pole on the unit circle, and with a power series that I can find. This should remind of the example $$\frac{1}{1-z} = \sum_{n=0}^\infty z^n$$ Here the coefficients are all equal to $1$, so they are bounded. More generally, we can place a simple pole at any point $a$ on the unit circle, and give it the residue of $b$, like this: $$\frac{b}{z-a}=\frac{-\bar a b}{1-\bar az} = -\bar a b \sum_{n=0}^\infty \bar a^n z^n$$ Again, the coefficients are bounded. And if we form a finite sum of such functions, the coefficients will be bounded still...
This sparks an idea (spoilered):