Prove that $\sum_{n=1}^\infty\frac{\sin(nz)}{2^n}$ is analytic on $\{z\in\mathbb{C}:|\operatorname{Im}(z)|<\log(2)\}$

Question

Prove that $f(z)=\sum_{n=1}^\infty\frac{\sin(nz)}{2^n}$ is analytic on $A=\{z\in\mathbb{C}:|\operatorname{Im}(z)|<\log(2)\}$

I tried expanding $\sin(nz)$ in terms of $e^{inz}$ but that did not help me unless I am doing something wrong. I know Weierstrass's M-test comes in to play.

Answer 1

Using $\sin nz=\frac{1}{2i}(e^{inz}-e^{-inz})$ works. The constant is irrelevant for the convergence.

Deal with the two exponentials separately. Let $z=x+iy$. Then $|e^{inz}|=e^{-ny}$, and $|e^{i((n+1)z}|=e^{-(n+1)y}$.

Thus, remembering about the $2^n$ in the denominator, we see that the norm of the ratio of two consecutive terms is $\frac{e^{-y}}{2}$. This norm is $\lt 1$ precisely if $e^{-y} \lt 2$, that is, if $y\gt -\log 2$.

In the same way, for the term in $e^{-iz}$, the norm of the ratio of two consecutive terms is $\lt 1$ precisely if $y \lt \log 2$. Thus, by the Weierstrass $M$-test, we have analyticity if $-\log 2\lt y\lt \log 2$.

Answer 2

We have $$ f(z)=\frac{1}{2i}\sum_{n=0}^\infty\frac{e^{inz}-e^{-inz}}{2^n} =\frac{1}{2i}\sum_{n=0}^\infty\left[\left(\frac{e^{iz}}{2}\right)^n-\left(\frac{e^{-iz}}{2}\right)^n\right] $$ If $|e^{\pm iz}/2|=\frac{1}{2}e^{\mp\text{Im}(z)}<1$, i.e. $\mp\text{Im}(z)<\log2$ or $|\text{Im}(z)|<\log2$ then the series converges and \begin{eqnarray} f(z)&=&\frac{1}{2i}\left[\frac{e^{iz}}{2}\frac{1}{1-e^{iz}/2}-\frac{e^{-iz}}{2}\frac{1}{1-e^{-iz}/2}\right]\cr &=&\frac{1}{2i}\left(\frac{e^{iz}}{2-e^{iz}}-\frac{e^{-iz}}{2-e^{-iz}}\right)\cr &=&\frac{2e^{iz}-1-2e^{-iz}+1}{2i(4-2e^{iz}-2e^{-iz}+1)}\cr &=&\frac{4i\sin z}{2i(5-4\cos z)}=\frac{2\sin z}{5-4\cos z}. \end{eqnarray} This function is obviously analytic!

Answer 3

Fix $\epsilon>0$.

If $|\Im(z)|<\log(2-\epsilon)$, then $$|\sin(nz)|=\left |\frac{e^{inz}-e^{-inz}}{2i}\right |\leq\frac{1}{2}\left (|e^{inz}|+|e^{-inz}|\right )=\frac{1}{2}\left (e^{-n\Im(z)}+e^{n\Im(z)}\right )<(2-\epsilon)^n,$$

so that the series converges uniformly on $A_\epsilon=\{z\in\mathbb{C}:|\text{Im}(z)|<\log(2-\epsilon)\}$ by the M-test.

Since the result of differentiating the series in the definition of $f$ term-by-term is $\sum_{n=1}^\infty\frac{n\sin(nz)}{2^n}$ which converges absolutely on $A_\epsilon$, it follows that differentiating term-by-term is legitimate and that $f$ is complex differentiable on $A_\epsilon$.

Since $\epsilon$ was arbitrary, we conclude that $f$ is complex differentiable on $A$. In particular, $f$ is holomorphic and hence analytic on $A$.