How can I solve $e^{-z}=2\mathrm{i}$ for complex $z$?

Question

Please, how can I solve $e^{-z}=2\mathrm{i}$ for complex $z$? Thank you

Answer 1

The solutions $z$ of the equation

$e^z=w$

are given by

$z= \log|w|+iarg(w) +2 k \pi i$, ($k \in \mathbb Z$).

Hence the solutions of $e^{-z}=2i$ are given by

$-z= \log|2i|+iarg(2i) +2 k \pi i$, ($k \in \mathbb Z$).

Can you proceed ?

Answer 2

Well, notice that in general for $\text{z}\in\mathbb{C}$:

$$\exp\left(-\text{z}\right)=\exp\left(-\left(\Re\left(\text{z}\right)+\Im\left(\text{z}\right)\cdot i\right)\right)=\exp\left(-\Re\left(\text{z}\right)-\Im\left(\text{z}\right)\cdot i\right)=$$ $$\exp\left(-\Re\left(\text{z}\right)\right)\cdot\exp\left(-\Im\left(\text{z}\right)\cdot i\right)=\exp\left(-\Re\left(\text{z}\right)\right)\cdot\left(\cos\left(\Im\left(\text{z}\right)\right)-\sin\left(\Im\left(\text{z}\right)\right)\cdot i\right)=$$ $$\exp\left(-\Re\left(\text{z}\right)\right)\cdot\cos\left(\Im\left(\text{z}\right)\right)-\exp\left(-\Re\left(\text{z}\right)\right)\cdot\sin\left(\Im\left(\text{z}\right)\right)\cdot i\tag1$$

So, when we have that (where $\text{a}\space\wedge\space\text{b}\in\mathbb{R}$):

$$\exp\left(-\text{z}\right)=\text{a}+\text{b}\cdot i\tag2$$

We need to solve the following system of equations:

$$ \begin{cases} \text{a}=\exp\left(-\Re\left(\text{z}\right)\right)\cdot\cos\left(\Im\left(\text{z}\right)\right)\\ \\ \text{b}=-\exp\left(-\Re\left(\text{z}\right)\right)\cdot\sin\left(\Im\left(\text{z}\right)\right) \end{cases}\tag3 $$

So, we can write:

$$\cos^2\left(\Im\left(\text{z}\right)\right)+\sin^2\left(\Im\left(\text{z}\right)\right)=\left(\frac{\text{a}}{\exp\left(-\Re\left(\text{z}\right)\right)}\right)^2+\left(\frac{\text{b}}{-\exp\left(-\Re\left(\text{z}\right)\right)}\right)^2=$$ $$\exp\left(2\cdot\Re\left(\text{z}\right)\right)\cdot\left(\text{a}^2+\text{b}^2\right)=1\space\Longleftrightarrow\space\Re\left(\text{z}\right)=\frac{1}{2}\cdot\ln\left(\frac{1}{\text{a}^2+\text{b}^2}\right)\tag4$$