Find all pairs$(a,b)$ of real numbers such that $(a+bi)^5=b+ai$

Question

Find all pairs$(a,b)$ of real numbers such that $(a+bi)^5=b+ai$

My Attempt,

If we let $a+bi=r(\cos\theta+ i \sin\theta)$. Then how should I move on. Should I have to do $(a+bi)^5=[r(\cos\theta +i \sin\theta)]^5$.

Please help. Thanks in advance

Answer 1

$z = a+bi\\ \bar z = a-bi\\ i\bar z = b + ai\\ z^5 = i \bar z\\ z = 0 \text { or}\\ z^6 = i\\ z^6 = e^{(2n + \frac 1{2})i}\\ z = e^{\frac {(4n+1)\pi}{12}i}$

Update $z = 0$ or

$z^6 = i$

De Moivre's theoem:

if $z = r(\cos \theta + i \sin\theta)$ then $z^n = r^n(\cos \theta + i \sin \theta)^n =r^n(\cos n\theta + i \sin n\theta)$

It also means that $z^{\frac 1n} = r^{\frac 1n}(\cos \theta + i \sin \theta)^{\frac 1n} =r^{\frac 1n}(\cos {\frac 1n}\theta + i \sin {\frac 1n}\theta)$

so here we have

$z^6 = i = (\cos (\frac \pi 2 + 2n\pi) + i \sin(\frac \pi 2 + 2n\pi))$

$z = $$(\cos (\frac \pi 2 + 2n\pi) + i \sin(\frac \pi 2 + 2n\pi))^{\frac 16}\\ (\cos (\frac \pi {12} + \frac {n\pi}{3}) + i \sin(\frac \pi {12} + \frac {n\pi}{3}))$

There are 6 solutions + the trivial solution.