Solving $z^4=(2+3i)^4$ To solve the equation, I calculated right side:
$z^4=(2+3i)^4=(-5+12i)^2=-119-120 i$
And then I get the correct answer:
$z_k=\underbrace{\sqrt[8]{119^2+120^2}}_{\sqrt{13}} \times Cis(\cfrac{\pi+\tan^{-1}(\frac{120}{119})}{4}+\cfrac{k \pi}{2}), k=0,1,2,3$
But, I am looking for a way to solve the equation $z^4=(2+3i)^4$ without expanding the right side. So I tried :
$z={ \left| r \right| }e^{i \theta}$
$r^4e^{4 \theta i}=(\sqrt{13} e^{(2k\pi+\tan ^{-1}(\frac{3}{2}))i})^4$
$r=\sqrt{13}$
$4\theta=4 \times {(2k\pi+\tan ^{-1}(\frac{3}{2}))}$
$\theta=2k\pi+\tan ^{-1}(\frac{3}{2})$
But I calculated the value of $\theta$ wrongly. How can I fix it?
 A: If $z^4=(2+3i)^4$ then $Z^4 = 1$ where $Z = \frac{z}{2+3i}$.
Hence the solutions set is
$$\{(2+3i), -(2+3i), i(2+3i), -i(2+3i)\}=\\ \{\sqrt{13} e^{i \phi},\sqrt{13} e^{i (\phi + \pi)},\sqrt{13} e^{i (\phi + \pi/2)},\sqrt{13}e^{i (\phi - \pi/2)}\}$$
where $\phi$ is such that $\cos \phi = \frac{2}{\sqrt{13}}, \sin \phi =\frac{3}{\sqrt{13}}$.
A: Alternatively, solve
$$
\left(\dfrac{z}{2+3i}\right)^4=1
$$
A: I would suggest you to go through this answer of mine.
Now... proceeding as above, We have one solution of the equation as $z=2+3i$ Just complete the square as the value of $n$ is $4$ here.
So your square looks something like this:

So those are your 4 solutions. :)
A: We have that
$$w^4=1 \iff w_k=i^k \quad k=0,1,2,3$$
then $(z\cdot w_k)^4=z^4$ and
$$z^4=(2+3i)^4 \iff z_k=(2+3i)\cdot i^k\quad k=0,1,2,3$$
A: Hint:
Use the fact
$$x^2-a^2=(x-a)(x+a)$$
and
$$x^2+a^2=(x-ai)(x+ai)$$
so
$$z^4-(2+3i)^4=0$$
$$\left ( x^2-(2+3i)^2 \right )\left ( x^2+(2+3i)^2 \right )=0$$
$$\left ( x-(2+3i) \right )\left ( x+(2+3i) \right )\left ( x-(2+3i)i \right )\left ( x+(2+3i)i \right )=0$$
A: one solution is obvious $$z_1=2+3i$$ the rest are distributed over a $90^{\circ}$ degree circle so: $$z_2=(2+3i)\cdot i$$ $$z_3=(2+3i)\cdot i\cdot i$$ $$z_3=(2+3i)\cdot i\cdot i\cdot i$$
A: Much simpler: $z^4 =(2+3i)^4= 1\cdot (2+3i)^4$
and $z = 1^{\frac 14} (2+3i)$, where $1^{\frac 14}$ is understood to mean the four complex fourth roots of $1$, namely $\pm 1, \pm i$.
A: It is a basic theorem that, once you have an $n$-th root $z$ of a complex  number, you obtain all its $n$-th roots multiplying $z$ by all the $n$-th roots of unity. What are the fourth roots of $1$?
