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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?

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    $\begingroup$ $z^4=(iz)^4=(-z)^4=(-iz)^4$, so $\pm(2+3i)$, $\pm i(2+3i)$ give you all roots of this $4$-th degree polynomial. $\endgroup$ – Sil Sep 24 '20 at 8:10
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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 $\cos \phi = \frac{2}{\sqrt{13}}, \sin \phi =\frac{3}{\sqrt{13}}$.

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  • $\begingroup$ Is there any way to write the polar form of the answers? I mean in the polar form as I wrote in the question we have $\tan ^ {-1} (\frac{120}{119}) $ ! where this come from? $\endgroup$ – Aligator Sep 18 '20 at 13:20
  • $\begingroup$ Yes, see updated answer. $\endgroup$ – mathcounterexamples.net Sep 18 '20 at 13:29
  • $\begingroup$ Thanks, but I still don't understand how these 4 answers are equivalent to: $$z_k=\sqrt{13} \times Cis(\cfrac{\pi+\tan^{-1}(\frac{120}{119})}{4}+\cfrac{k \pi}{2}), k=0,1,2,3$$ $\endgroup$ – Aligator Sep 18 '20 at 13:38
  • $\begingroup$ What is $Cis$ ? $\endgroup$ – mathcounterexamples.net Sep 18 '20 at 13:48
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    $\begingroup$ This is coming from the fact that $\cfrac{\pi+\tan^{-1}(\frac{120}{119})}{4} = \tan^{-1}(\frac{3}{2}) \approx 0.982793723247329$. $\endgroup$ – mathcounterexamples.net Sep 18 '20 at 13:57
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Alternatively, solve $$ \left(\dfrac{z}{2+3i}\right)^4=1 $$

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  • $\begingroup$ I write: $\left(\dfrac{z \times(2-3i)}{\sqrt{13}}\right)^4=1$ and $z^4 \times(2-3i)^4=169$ and I should expand the expression again. $\endgroup$ – Aligator Sep 18 '20 at 12:51
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    $\begingroup$ Soheil: No. Stop expanding. You're making unnecessary work for yourself. Look at the answer by @mathcounterexamples to see where to go next. $\endgroup$ – JonathanZ supports MonicaC Sep 18 '20 at 13:00
  • $\begingroup$ Oh I get it thanks. $\endgroup$ – Aligator Sep 18 '20 at 13:05
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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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    $\begingroup$ i like your atitude,You have not asked any questions, But you continue to keep answering and helping people $\endgroup$ – Albus Dumbledore Sep 18 '20 at 13:18
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    $\begingroup$ A lot of thanks. I was looking forward to it. That encourages me to do better. Thanks again :) $\endgroup$ – Soumyadwip Chanda Sep 18 '20 at 13:32
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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

$$z^4=(2+3i)^4 \iff z_k=(2+3i)\cdot i^k\quad k=0,1,2,3$$

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  • $\begingroup$ @lhf Do you mean taking $w=i$? $\endgroup$ – user Sep 18 '20 at 12:55
  • $\begingroup$ @lhf Thanks for the suggestion! I put in this way. $\endgroup$ – user Sep 18 '20 at 12:56
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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 $\pm 1, \pm i$.

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Hint: Use the fact $$x^2-a^2=(x-a)(x+a)$$ and $$x^2+a^2=(x-ai)(x+ai)$$ so $$z^4-(2+3i)^4=0$$ $$\left ( x^2-(2+3i)^2 \right )\left ( x^2+(2+3i)^2 \right )=0$$ $$\left ( x-(2+3i) \right )\left ( x+(2+3i) \right )\left ( x-(2+3i)i \right )\left ( x+(2+3i)i \right )=0$$

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one solution is obvious $$z_1=2+3i$$ the rest are distributed over a $90^{\circ}$ degree circle so: $$z_2=(2+3i)\cdot i$$ $$z_3=(2+3i)\cdot i\cdot i$$ $$z_3=(2+3i)\cdot i\cdot i\cdot i$$

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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 $1$?

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