# Find the radius of convergence of the complex power series.

Find the radius of convergence of the power series $$\displaystyle \sum_{n=0}^{\infty} \left(\int_{|z|=1} \frac{\cos{\left(\frac{1}{z-10}\right)}}{z^{n+1}} \,dz\right) z^n$$.

I have no idea how to approach this. I tried to solve the integral using the fact that $$\int_{\gamma}f(z)dz=\int_a^b f(\gamma(t)) \gamma'(t)dt$$ and then applied the ratio test but didn't succeed.

Also, since $$z \in \mathbb{C}$$, we have that $$\cos{\left(\frac{1}{z-10}\right)}$$ is unbounded by the Liouville's Theorem.

Any help would be appreciated.

• Hint: ignore the specific function for the moment and consider the general expression $$\sum_{n=0}^{\infty} \left(\int_{|z|=1} \frac{f(z)}{z^{n+1}} \,dz\right) z^n.$$ Can you use the Cauchy integral formula to simplify the integral? Do you recognize the series that results? Commented Jan 15, 2020 at 19:36
• Let me call the variable of the power series $w$, to not confuse it with the variable of integration $z$. On compacts inside the disc of convergence you can swap the integral and sum to get $\int_{|z|=1}\frac{1}{z}\frac{\cos(1/(z-10))}{1-w/z}dz=\int_{|z|=1}\frac{\cos(1/(z-10))}{z-w}dz$. Then you can apply Cauchy's formula. Commented Jan 15, 2020 at 20:31

Sketch: Let $$f(z)= \cos (1/(z-10)).$$ Note that $$f$$ is holomorphic in $$\{|z|<10\}$$ but has a naughty singularity at $$10.$$ It follows that the power series of $$f,$$ centered at $$0,$$ has radius of convergence exactly equal to $$10.$$
On the other hand, the power series of $$f,$$ centered at $$0,$$ equals $$\sum_{n=0}^{\infty}\dfrac{f^{(n)}(0)}{n!}z^n.$$ There is a well known formula for $$\dfrac{f^{(n)}(0)}{n!}$$ ...
• Why does the power series of $f$, centered at $0$, have radius of convergence equal to $10$? Commented Jan 15, 2020 at 22:43
• Because if $f$ is holomorphic in $D(0,r)$ then $f$ equals its Taylor series, centered at $0$ in the same disc. So the ROC is at least $r.$ But if $f$ has a singularity on the boundary, then the radius of convergence is at most $r.$