Solve complex equation with exponential I have to solve:
$$e^{3z}+3ie^{2z}-ie^z+3=0$$
My attempt:
Let $0\ne x:=e^z$. Then we can rewrite our equation as:
$$x^3+3ix^2-ix+3=0$$
$$ix^2(-ix+3)+(-ix+3)=0$$
$$(-ix+3)(ix^2+1)=0$$
So $x\in \{-3i,\sqrt{i},-\sqrt{i}\}$
Now going back to our subtitution we have:
$$e^z=-3i \lor e^z=\sqrt{i} \lor e^z=-\sqrt{i}$$
In the first case: $z=\ln3+i(\frac{-\pi}{2}+2k\pi)$ where $k\in\mathbb{Z}$
In the second case: $z=i(\frac{\pi}{4}+2l\pi)$ where $l\in\mathbb{Z}$
In the third case: $z=i(\frac{- 3\pi}{4}+2m\pi)$ where $m\in\mathbb{Z}$
So those above are solutions for our initial equation?
 A: $$e^{3z}+3ie^{2z}-ie^z+3=0$$
Gives:
$$e^{3z}+((3i)(e^{2z}))-((i)(e^z))+3e^{0i}=$$
$$e^{3z}+((3e^{\frac{1}{2}\pi})(e^{2z}))-((i)(e^z))+3e^{0i}$$
It gives you the following equation:
$$(e^z+3i)(e^{2z}-i)=0$$
So we know that or the first part is zero or the second part!
$$(e^z+3i)=0$$
$$e^z=-3i$$
$$z=\frac{1}{2}i(4\pi n-\pi -2i*ln(3))$$
With n is the element of Z (set of integers)!
Or:
$$(e^{2z}-i)=0$$
$$e^{2z}=i$$
$$z=\frac{1}{4}i(4\pi n+\pi )$$
With n is the element of Z (set of integers)!
A: Remember that $\forall z \in \mathbb C, \forall z' \in \mathbb C, e^z=e^{z'}\iff \exists k \in \mathbb Z: z=z'+2ik\pi \color{red}{(*)}$
Here is the method to solve this type of drill, I believe this is what OP needs to be sure; it could be explained with letters but I prefer to do it with the numbers of the exercise, which cover all the cases:
$i=e^{i\frac{\pi}{2}}=e^ {2i\frac{\pi}{4}}=(e^ {i\frac{\pi}{4}})^2$. So, $(e^z)^2=i \iff (e^z)^2-(e^ {i\frac{\pi}{4}})^2=0 $
And we then find the second case and third case, according to $\color{red}{(*)}$.
$-1=e^{i\pi}$. So, $-3i=e^{-i\frac{\pi}{2}}e^{log3}=e^{log3-i\frac{\pi}{2}}$.
And we then find the first case, according to $\color{red}{(*)}$.
