Solving $z^4=(2+3i)^4$

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 $\cos \phi = \frac{2}{\sqrt{13}}, \sin \phi =\frac{3}{\sqrt{13}}$.


Alternatively, solve $$ \left(\dfrac{z}{2+3i}\right)^4=1 $$


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:

enter image description here

So those are your 4 solutions. :)

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Complex Numbers

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