How to find all the $z$ that satisfy $(1+i)z^4=(1-i)|z|^2$?

Question

Would you please help me solve this? I need all the $z$ that satisfy the equality

$$(1+i)z^4=(1-i)|z|^2.$$

I tried doing this:

$$ \begin{aligned} (1+i)z^4&= (1-i)z\overline z\\ (1+i)z^4 -(1-i)z\overline z &= 0\\ z[(1+i)z^3-(1-i)\overline z]&= 0 \end{aligned} $$

Then $z= 0$ and $$(1+i)z^3-(1-i)\overline z= 0.$$

I don't know what to do with $\overline z$.

Answer 1

Hint

$$(1+i)z^4=(1-i)|z|^2\iff z^4=\dfrac{1-i}{1+i}|z|^2=-i|z|^2.$$

Taking modulus we have $|z|=0$ or $|z|=1.$ In the first case it is $z=0.$ In the second case $z=e^{i\theta}.$ So, we have

$$\cos (4\theta)+i\sin(4\theta)=-i.$$ That is, we have to solve

\begin{cases}\cos(4\theta)=0\\\sin(4\theta)=-1\end{cases}

Answer 2

$$(1+i)z^4=(1-i)|z|^2\iff iz^4=z\bar z\iff\begin{cases}z=0\\iz^3=\bar z\end{cases}$$ $$iz^3=\bar z\iff i(a+ib)^3=(a-ib)\iff {ia^3 - 3 a^2 b -i 3 a b^2 + b^3}=(a-ib)\\\iff b^3- 3 a^2 b +i(a^3 - 3 a b^2)=a-ib\iff\begin{cases}b^3- 3 a^2 b=a\\a^3 - 3 a b^2=-b\end{cases}$$

We can also notice that $|b^3- 3 a^2 b +i(a^3 - 3 a b^2)|=\sqrt{(a^2+b^2)^3}=\sqrt{a^2+b^2}=|a-ib|\implies |z|=\begin{cases}1\\0\end{cases}$, so it comes with terms with @mfl's answer