Let $ z\in \mathbb{C} $ and $\sin(z)=0$. Prove that $z\in \mathbb{R}$ Assume that $$ z\in \mathbb{C},\ \sin(z)=0 $$ and follow that $$ z\in \mathbb{R}. $$
What I have tried so far: I went for comparing with the Euler's form. Since we can say that for all z $$ \sin(z)=\frac{e^{iz}-e^{-iz}}{2i}, \ \cos(z)=\frac{e^{iz}+e^{-iz}}{2} $$ we can follow that $$ e^{iz}=\cos(z)+i\sin(z). $$ so, if we let sin(z) be zero, conforming to the task, we can transform $$ e^{iz}=\cos(z)+i*0 = \cos(z). $$ and since $$ Re(e^{iz}) = \cos(z), \ \cos(z)\in \mathbb{R}$$ we can say that also $$ z\in \mathbb{R} $$ is this a valid way of showing the argument, or would I have to use the following as a vantage point: $$ z= r e^{i\phi}, \ for \ z=x+iy $$ and somehow transform that term to get the solution. I don't really see how I would get the sin(z)=0 to aid me when it's not in the Euler form of the $r e^{i\phi}$ form. Any help would be super appreciated.
Edit: Thanks a lot for the replies, everyone. I'm sorry for my onesided thinking since I'm still a calc 1 newbie. I'll take a look and re-evaluate.
 A: $$\sin(z)=\frac{e^{iz}-e^{-iz}}{2i}=0$$implies$$e^{iz}=e^{-iz}$$
Use $$z=x+iy$$ with $x,y\in\mathbb R$
Then $$e^{i(x+iy)}=e^{-i(x+iy)}\\e^{ix}e^{-y}=e^{-ix}e^{y}$$
Now take the absolute value, and notice that $|e^{ix}|=1$:
$$|e^{ix}e^{-y}|=|e^{-ix}e^{y}|\\|e^{ix}||e^{-y}|=|e^{-ix}||e^{y}|\\|e^{-y}|=|e^y|$$
Since $y\in \mathbb R$ it means that $$e^y=\frac1{e^y}=1$$or $y=0$.
A: $0 = \sin(z) = (e^{iz} - e^{-iz})/(2i) \iff e^{iz} = e^{-iz} \iff e^{2iz} = 1$
Write $z$ as $a + bi$
$1e^{0} = 1 = e^{2iz} = e^{2i(a + bi)} = e^{-2b+2ai} = e^{-2b} e^{2ai}$
Equating lengths and angles we have:
$1 = e^{-2b} \implies b = 0$
$0 = 2a + 2 \pi n \implies a = \pi k$
So $z = \pi k$ - all of these are real
A: We know that $sin(z) = \frac{e^{iz} - e^{-iz}}{2i} = 0$. Put $t = e^{iz}$, then $t = 1/t$, $t^2 = 1$, $t = \pm 1$. Hence $e^{iz} = \pm 1$.
As $z = a+bi$ where $a,b \in \mathbb{R}$, we have $e^{iz} = e^{-b+ia} = \pm1$.
Thus $|e^{-b+ia}| = |\pm 1| = 1$, but $|e^{-b+ia}| = |e^{-b}| \cdot |e^{ia}| = e^{-b} \cdot 1$. Hence $e^{-b} = |e^{-b+ia}|  = 1$ and $b=0$. That means that $z = a+bi$ and $b=0$, q.e.d.
A: First, we have
$$\sin(a + bi) = \sin(a)\cos(bi) + \cos(a)\sin(bi).$$
Then
$$\sin(bi) = \frac{e^{-b}-e^{b}}{2i} = \sinh(b)i, \text{ and } \cos(bi) = \frac{e^{-b}+e^b}{2}=\cosh(b).$$
So we get
$$\sin(a + bi) = \sin(a)\cosh(b)+\cos(a)\sinh(b)i.$$
If this is a real number then either $\cos(a) = 0$ or $\sinh(b) = 0 \iff b = 0$. Maybe you can work out the rest from here?
