# 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?

• $z^4=(iz)^4=(-z)^4=(-iz)^4$, so $\pm(2+3i)$, $\pm i(2+3i)$ give you all roots of this $4$-th degree polynomial. – Sil Sep 24 '20 at 8:10

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

• 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? – Aligator Sep 18 '20 at 13:20
• Yes, see updated answer. – mathcounterexamples.net Sep 18 '20 at 13:29
• Thanks, but I still don't understand how these 4 answers are equivalent to: $$z_k=\sqrt{13} \times Cis(\cfrac{\pi+\tan^{-1}(\frac{120}{119})}{4}+\cfrac{k \pi}{2}), k=0,1,2,3$$ – Aligator Sep 18 '20 at 13:38
• What is $Cis$ ? – mathcounterexamples.net Sep 18 '20 at 13:48
• This is coming from the fact that $\cfrac{\pi+\tan^{-1}(\frac{120}{119})}{4} = \tan^{-1}(\frac{3}{2}) \approx 0.982793723247329$. – mathcounterexamples.net Sep 18 '20 at 13:57

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

• 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. – Aligator Sep 18 '20 at 12:51
• Soheil: No. Stop expanding. You're making unnecessary work for yourself. Look at the answer by @mathcounterexamples to see where to go next. – JonathanZ supports MonicaC Sep 18 '20 at 13:00
• Oh I get it thanks. – Aligator Sep 18 '20 at 13:05

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. :)

• i like your atitude,You have not asked any questions, But you continue to keep answering and helping people – Albus Dumbledore Sep 18 '20 at 13:18
• A lot of thanks. I was looking forward to it. That encourages me to do better. Thanks again :) – Soumyadwip Chanda Sep 18 '20 at 13:32

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

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

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

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 $$\pm 1, \pm i$$.

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

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

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 $$1$$?