Prove that the taylor series of cos(z) and sin(z) are holomorphic 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!
 A: It is easy to see that $\cos(z) = (e^{iz} + e^{-iz})/2$, either by looking at the power series, or from $e^{iz} = \cos(z) + i\sin(z)$.  (In fact, in some courses the cosine function is defined this way!)  Since both exponentials are entire (i.e. holomorphic in the whole complex plane), their sum is also.  For $\sin(z)$, replace the $+$ sign with a $-$ sign and divide by $2i$ instead of $2$.
Of course, this assumes that you already know the exponential function is entire.  If not, you can either attack the sine and cosine series directly, or just prove that the exponential series gives an entire function and then use the above argument.
One way would be to recall that if the ratio of successive coefficients $a_{n+1}/a_n$ has a limit $L$, the radius of convergence of the power series is $1/L$.  Since the ratio is $i/(n+1)$, its limit as $n\rightarrow\infty$ is zero, and the radius of convergence is infinite.
A: a) The ratio test works well
b) The power series converge locally uniformly, so the sum is holomorphic. (Something similar to this is almost certainly a theorem in your textbook, i.e. the sum of a convergent power series is holomorphic.)
