How to solve the equation $\exp(iz)=-e$ How do I solve the equation 

$$\exp(iz)=-e$$

Can anyone please explain the procedure to solve this kind of question to me please?

Much appreatiate
Solution with comments:  $e^{iz -1} = -e$:

1.)  $e^{iz} = -e \Rightarrow e^{iz -1} = -1= e^{2\pi ni}e^{\pi i} = e^{(2n + 1) \pi i}$, where $n \in \Bbb Z$;

2.)  $e^{iz - 1} =  e^{(2n + 1) \pi i} \Rightarrow e^{iz - 1 - (2n + 1)\pi i} = 1$;

3.)  $e^{iz - 1 - (2n + 1)\pi i} = 1 \Rightarrow iz - 1 - (2n + 1)\pi i = 2m \pi i$, where $m \in \Bbb Z$;

4.) $iz - 1 - (2n + 1)\pi i = 2m \pi i \Rightarrow iz - 1 - 2(m + n + 1) \pi i = 0$
$\Rightarrow iz - 1 - (2k + 1) \pi i = 0,$ where $k = m + n \in \Bbb Z$;

5.) $iz - 1 - (2\pi k + 1) i = 0 \Rightarrow iz = 1 + (2\pi k + 1)  i \Rightarrow z = -i + (2k + 1) \pi$.

We check the solutions $z = -i + (2k + 1) \pi$ obtained in (5):

A.)  $z = -i + (2k + 1) \pi \Rightarrow iz = 1 + (2k + 1) \pi i \Rightarrow iz - 1 = (2k + 1)\pi i$;

B.)  $iz -1 = (2k + 1) \pi i \Rightarrow e^{iz - 1} = e^{(2k + 1) \pi i} = e^{2k \pi i} e^{\pi i} = -1$.

The above shows that all solutions of

$e^{iz - 1} = -e \tag{6}$

are of the form

$z = -i + (2k + 1)\pi, \tag{7}$

with $k \in \Bbb Z$, and that every $z$ as in (7) solves (6).

Note that we have directly used the fact that $e^w = 1 \Leftrightarrow w = 2 k \pi i, \;\; k \in \Bbb Z$, instead of invoking the logarithm function(s), $\ln$ or $\text{Ln}$, in an appropriate open set of $1$ in the punctured complex plane $\Bbb C \setminus \{0\}$.  This fact is elementary and easy to prove by writing $z = x + iy$ and then $e^z = e^x(\cos y + i \sin y) = 1$, and avoids the subtleties of the complex logarithm.  For these reasons, I chose this line of attack for my solution.

Hope this helps.  Cheers,

and as always,

Fiat Lux!!!
$e^{iz} = -e$

$e^{iz} = e^{\pi i}e$

$e^{iz} = e^{\pi i + 1}$

$ln(e^{iz}) = ln(e^{\pi i + 1})$

$iz\ \ln(e) = (\pi i + 1)ln(e)$

Can you take it from here?