So I need to find the power series of $\left(\frac{\sin z}z\right)^2$ around $0$, and find its radius of convergence.
Also the function is defined to be $1$ in $z=0$.
My guess about the radius is infinity, but I'm stuck in the power series.
This is what I've done so far:
$$\left(\frac{\sin z}z\right)^2 = \left(\frac{z-\frac{z^3}{3!}+\frac{z^5}{5!}-\cdots\frac{z^{2n+1}}{(2n+1)!}+\cdots}z\right)^2$$ $$=\left(1-\frac{z^2}{3!}+\frac{z^4}{5!}-\cdots\frac{z^{2n}}{(2n+1)!}+\cdots\right)^2$$ $$=\left(\sum_{n=0}^{\infty}\frac{(-1)^nz^{2n}}{(2n+1)!}\right)^2$$
could really use some help. Thanks very much!
Using Lord Shark the Unknown's hint we can rewrite the problem in the following way
$$\left(\frac{\sin z}z\right)^2=\frac1{2z^2}(1-\cos 2z)$$
Utilizing the well known Taylor Series of the cosine function this further becomes
$$\begin{align} \frac1{2z^2}(1-\cos 2z)&=\frac1{2z^2}\left(1-\sum_{n=0}^{\infty}(-1)^n\frac{(2z)^{2n}}{(2n)!}\right)\\ &=\frac1{2z^2}-2\sum_{n=0}^{\infty}(-1)^n\frac{(2z)^{2n-2}}{(2n)!}\\ &=-2\sum_{n=1}^{\infty}(-1)^n\frac{(2z)^{2n-2}}{(2n)!} \end{align}$$
Regarding the radius of convergence one could apply the ratio test to verfity that it is actually infinity.