I have a question on an old exam paper given as follows:

> 
> If we define
> 
> 
> $$\cos(z) = 1 - \frac{z^2}{2!} + \frac{z^4}{4!} - ... \frac{z^{2n}}{(2n)!} + ... = \sum\_{n=0}^{\infty} \frac{(-1)^nz^{2n}}{(2n)!}$$
> 
> 
> $$\sin(z) = z - \frac{z^3}{3!} + \frac{z^5}{5!} - ... \frac{z^{2n+1}}{(2n+1)!} + ... = \sum\_{n=0}^{\infty} \frac{(-1)^nz^{2n+1}}{(2n+1)!}$$
> 
> 
> Then:
> 
> 
> a) Prove that both series converge in the whole complex plane.
> 
> 
> b) Prove that $\cos(z)$ and $\sin(z)$ are holomorphic functions in the whole complex plane.
> 
> 
> You can use without proof that the derivative of $z^n$ is $nz^{n-1}$ and the algebraic properties of derivatives hold.
> 
> 
> 

I believe I'm correct in part (a) by using D'Alembert's ratio test, but it's part (b) I don't understand. I've tried Cauchy-Riemann equations, but I can't easily separate the imaginary and real terms.

Any help on this would be most appreciated!