How to solve an equation with a complex exponential equalling a complex number

Question

I want to solve the following equation for $\theta$

$$ e^{i2\theta} = x $$

where both $x$ and $\theta$ can be complex numbers. Judging from WolframAlpha's solution it is not simply taking the logarithm and rearranging for $\theta$. What is the correct way to solve this?

Answer 1

Write $x = e^{\text{Log} x}$ ($\text{Log}$ is the principal branch of logarithm - defined for $x \neq 0$; note that for $x=0$ there isn't a solution), and note that:

$$e^z = e^w \iff z = w + 2ik\pi, \text{ for some $k \in \Bbb Z$}$$

Answer 2

Well, we know Euler's formula:

$$e^{\varphi i}=\cos\varphi+i\sin\varphi\tag1$$

So, when we have for your LHS:

$$e^{2\theta i}=e^{2\left(\sigma+\zeta i\right)i}=e^{2\sigma i-2\zeta}=e^{2\sigma i}\cdot e^{-2\zeta}=\frac{\cos\left(2\sigma\right)+\sin\left(2\sigma\right)i}{e^{2\zeta}}\tag2$$

Where $\sigma$ is the real part of $\theta$ and $\zeta$ is the imaginary part of $\theta$.

So, we can setup a system of equations:

$$ \begin{cases} \frac{\cos\left(2\sigma\right)}{e^{2\zeta}}=\Re\left(x\right)\\ \\ \frac{\sin\left(2\sigma\right)}{e^{2\zeta}}=\Im\left(x\right) \end{cases}\space\space\space\space\space\space\therefore\space\space\space\space\space\space\frac{\cos\left(2\sigma\right)}{\Re\left(x\right)}=\frac{\sin\left(2\sigma\right)}{\Im\left(x\right)}\space\Longleftrightarrow\space\tan\left(2\sigma\right)=\frac{\Im\left(x\right)}{\Re\left(x\right)}\tag3 $$

Answer 3

For given $c\ne0$ the equation $e^z=c$ has infinitely many solutions $z\in{\mathbb C}$. The solution set $S$ is given by $$S=\log|c|+ i\>{\rm arg}(c)$$ and is a one-dimensional lattice in the vertical direction, spaced by $2\pi i$. If $z\in S$ is written as $z=2i\theta$, and the given $c$ is named $x$, then this amounts to $$\theta={1\over2i}\log|x|+{1\over2}{\rm arg}(x)\ .$$