Solving $\sin(z)=0$ for complex numbers $z$ by using Euler's formula 
Using Euler's formula prove the following:
  $$\sin(z)=0\Leftrightarrow z=\pi k\text{ for } k\in\mathbb{Z}.$$

I don't even know where to start.
 A: Hint. Note that the sine with complex argument is defined as
$$\sin(z):=\frac{e^{iz}-e^{-iz}}{2i}=\frac{e^{i(x+iy)}-e^{-i(x+iy)}}{2i}=
\frac{e^{-y}e^{ix}-e^{y}e^{-ix}}{2i}$$
where $z=x+iy$ with $x,y\in\mathbb{R}. Then by using How to prove Euler's formula: $e^{i\varphi}=\cos(\varphi) +i\sin(\varphi)$?
we get
$$\sin(z)=0\Leftrightarrow e^{-y}(\cos(x)+i\sin(x))=e^{y}(\cos(x)-i\sin(x)) \Leftrightarrow
\begin{cases}
\sin(x)(e^y+e^{-y})=0 \\
\cos(x)(e^y-e^{-y})=0
\end{cases}$$
Can you take it from here?
A: It depends on want you want to prove and how you define things... 
A way to see this question is to ask if
$$ \forall z\in \mathbb{C}, \quad \sin(z)=0 \qquad \Longleftrightarrow \qquad  z=\pi k , \quad k\in \mathbb{z}$$ 
knowing that the same statement is true over $\mathbb{R}$... With $z=x+iy$
\begin{align*}
 \sin(z):= \frac{e^{iz}-e^{-iz}}{2i} =0 &\qquad \Longleftrightarrow \qquad e^{iz}=e^{-iz} \\
&\qquad \Longleftrightarrow \qquad e^{ix}e^{-y}=e^{-ix}e^{y}
\end{align*}
1) 
$e^{ix}e^{-y}=e^{-ix}e^{y} \Rightarrow |e^{ix}e^{-y}|=|e^{-ix}e^{y}| \Rightarrow e^{-y}=e^{y} \Rightarrow y=0$, since $t\mapsto e^t$ is strictly increasing.
2)With $y=0$,
\begin{align*}
e^{ix}e^{-y}=e^{-ix}e^{y} &\quad \Rightarrow \quad e^{ix} =e^{-ix} \\
&\quad \Rightarrow \quad \cos(x)+i\sin(x) =\cos(-x)+i\sin(-x) \\
&\quad \Rightarrow \quad \cos(x)+i\sin(x) =\cos(x)-i\sin(-x) \\
&\quad \Rightarrow \quad  \sin(x) =0 \\
&\quad \Rightarrow \quad  x=k\pi, \quad k\in \mathbb{Z} \\
\end{align*}
So $z=k\pi+i*0=k\pi$, the converse being obvious.
A: Euler's formula says that the complex exponential $e^{iz}$ can be expressed as a sum of sinusoidals:
$$e^{iz}=cos(z) + i\ sin(z)$$. Does that help?
