Solving Complex Valued Equalities I was thinking the following. Say I want to solve
$$
(i)^z=\sqrt 3 +2i.
$$
Usually, at least the way I have been taught, we write in polar form the other side of the inequality as
$$
(i)^z =e^{\pi/3+2\pi ik}
$$
and the left-hand side as
$$
(e^{\pi/2})^z =e^{\pi/3+2\pi ik}
$$
and then solve by expanding $z=x+iy$ in real and imaginary powers.
My (quick ) question: Don't we need to also consider $i$ as $i=e^{\pi/2+2\pi ni}$? When doing so I get more solutions.
Remark: I don't need someone to do the algebra for me, I just need someone to tell me what generality we need consider.
 A: First, if you are calculating $~\displaystyle A = \left[re^{i\theta}\right]^{x + iy},$ you normally adopt some sort of constraint on $\theta$ confining it to a half-open interval of width $2\pi$.  This is because you want $A$ to evaluate unambiguously, for each value of $z = (x + iy)$.
I will adopt the arbitrary convention that $-\pi < \theta \leq \pi.$
Then, $~\displaystyle i = [0] + i[1] = \cos(\pi/2) + i\sin(\pi/2) = e^{i\pi/2}.$
The value $z_1 = \sqrt{3} + 2i$ may be expressed as
$\displaystyle \sqrt{7} \times \left[\frac{\sqrt{3}}{\sqrt{7}} + i\frac{2}{\sqrt{7}}\right].$
Let $\alpha = \text{Arccos}\left(\frac{\sqrt{3}}{\sqrt{7}}\right).$
Then, $z_1$ may be expressed as $\displaystyle ~\sqrt{7}e^{i(\alpha + 2k\pi)} ~: ~k \in \Bbb{Z}.$

Edit
Note that $~\displaystyle \left(\frac{\sqrt{3}}{\sqrt{7}}\right)^2 + \left(\frac{2}{\sqrt{7}}\right)^2 = 1.$

This means that any value of $z = (x + iy)$ that results in
$$\left[e^{i(\pi/2)}\right]^{(x + iy)} = \sqrt{7}e^{i(\alpha + 2k\pi)} ~: ~k \in \Bbb{Z} \tag1 $$
must be construed to be a solution.
The constraint in (1) above may be equivalently expressed as
$$e^{-y\pi/2} \times e^{ix\pi/2} = e^{\log(\sqrt{7})} \times e^{i(\alpha + 2k\pi)} ~: ~k \in \Bbb{Z}. \tag2 $$
Therefore, the set of all satisfying values $(x + iy)$ is given by

*

*$\displaystyle -y\pi/2 = \log(\sqrt{7})$


*$\displaystyle x\pi/2 = \alpha + 2k\pi ~: ~k \in \Bbb{Z}.$
