I have 2 problems that I cannot finish/ not sure.

Solve for all values of z such that,

1) $\displaystyle sin(z) = 0$

$\displaystyle sin(z) = \frac{e^{iz}-e{-iz}}{2i} = 0$

$\displaystyle e^{iz} = e^{-iz}$

$\displaystyle \frac{e^{iz}}{e^{-iz}} = 1$

$\displaystyle e^{2iz} = 1$

$\displaystyle ln(e^{2iz}) = ln(1)$

$\displaystyle 2iz = 0$

$\displaystyle z = 0$

However, my problem is z can be $\displaystyle \pi$ or $\displaystyle k\pi$. What should I do?

2) $\displaystyle e^{iz} = -1$

$\displaystyle e^{iz} = i^2$

$\displaystyle ln(e^{iz}) = ln(i^2)$

$\displaystyle iz = 2ln(i)$

$\displaystyle z = \frac{2ln(i)}{i}$

From $\displaystyle z = \frac{2ln(i)}{i}$, is this a final answer?

Thank you.