I managed to show that: $$\sum_{z=1}^{\infty} \frac{sin(nz)}{2^n}$$ is analytic on $\{z \in \mathbb{C}| -log(2)<Im(Z)<log(2)\}$ I am assuming that is correct?
I am now stuck as to how to evaluate the series on that set. How would I get started?
I have so far re-written: $$\sum_{z=1}^{\infty} \frac{sin(nz)}{2^n} = \frac{1}{2i}[\sum \frac{e^{inz}}{2^n} - \sum \frac{e^{-inz}}{2^n}]$$
Thank you!
Let $z=x+iy$ where $x,y$ are real
Now both the infinite geometric Series will converge if
$$\left|\dfrac{e^{i(x+iy)}}2\right|,\left|\dfrac{e^{-i(x+iy)}}2\right|<1$$
$$\iff e^{-y}<2, e^y<2$$
$$\iff -y<\ln2, y<\ln2$$
If $y>0, \ln2>-y,$ so it is sufficient to have $y<\ln2$
If $y<0,\ln2>0>y,$ so it is sufficient to have $-y<\ln2\iff y>-\ln2$
So, we need $-\ln 2<y<\ln2$
In that case, $\displaystyle\sum_{n=1}^\infty\dfrac{e^{izn}}{2^n}=-1+\sum_{n=0}^\infty\left(\dfrac{e^{iz}}2\right)^n=-1+\dfrac1{1-\dfrac{e^{iz}}2}=-1+\dfrac2{2-e^{iz}}$
Similarly,$\displaystyle\sum_{n=1}^\infty\dfrac{e^{-izn}}{2^n}=-1+\dfrac2{2-e^{-iz}}$
$$2i\sum_{n=1}^\infty\dfrac{\sin(nz)}{2^n}=\dfrac2{2-e^{iz}}-\dfrac2{2-e^{-iz}}=\dfrac{4(e^{iz}-e^{-iz})}{2^2+1-2(e^{iz}+e^{-iz})}$$
Now use How to prove Euler's formula: $e^{i\varphi}=\cos(\varphi) +i\sin(\varphi)$?
