Radius of convergence of $\frac{1}{\cosh(z)}$

Question

I know (wrongly I found out!) that $\frac{1}{\cosh(z)}=1+\frac{2}{z^2}+\frac{24}{z^4}+...$ centered at the origin.

Does this mean that the radius of convergence is 0 since the only singularity is at the origin as well? So the distance from the origin to a singularity is 0?

I am confused about this. Thanks.

Answer 1

The radius of convergence is the distance in the complex plane to the nearest singularity. Now $\cosh (z) = 0$ when $z= \pm \pi \mathrm{i}/2$, so the radius of convergence is $\pi/2$.

Answer 2

Solve this equation in the complex numbers:

$$\cosh(z) = 0.$$

You will now know the radius of convergence of the Taylor expansion.

Answer 3

If you perform the expansion around $z=0$, you effectively have $$\cosh(z)=1+\frac{z^2}{2}+\frac{z^4}{24}+\frac{z^6}{720}+\frac{z^8}{40320}+O\left(z^{10}\right)$$ making $$\frac 1 {\cosh(z)}=\frac 1{1+\frac{z^2}{2}+\frac{z^4}{24}+\frac{z^6}{720}+\frac{z^8}{40320}+O\left(z^{10}\right) }$$ Using long division $$\frac 1 {\cosh(z)}=1-\frac{z^2}{2}+\frac{5 z^4}{24}-\frac{61 z^6}{720}+\frac{277 z^8}{8064}+O\left(z^{10}\right)$$