Term by term differentiation of sin z

Question

In complex numbers, $\sin\:z$ is expanded as the series \begin{align*} \sum_{n \geq 0} (-1)^n \frac{z^{2n+1}}{(2n+1)!} \end{align*} with pointwise convergence (right?)

Then, when we find an expansion of cosine by using term by term differentiation? I think that I should show that this series uniformly converges to $\sin\:z$. Then, How can I show that without using Taylor theorem ?

Answer 1

$$\sin z = \sum_{n \geq 0} (-1)^n \frac{1}{2n+1}z^{2n+1}.$$

Theorem. Let $f_n$ be sequence analytic on domain $D$. If $f_n$ converges uniformly on every compact subset of $D$ , then $f_n$ analytic on $D$.

Property. If powerseries centered at $z_0$ has radius R of converge, then the powerseries converes uniformly on everycompact subset lying in domain where power series converges.

So, $\sum_{n \geq 0} (-1)^n \frac{1}{2n+1}z^{2n+1}.$ is analytic on complex plane.

Theorem. $f_n$ is sequence in domain D. If $\sum f_n$ converges in D and converges uniformly on every compact subset of D, then $\sum f_n$ is analytic on D and its derivative is $\sum f_n ^\prime$

You can get result that you have wanted.