mjhjmtu

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:

$$
beginaligned
(1+i)z^4&= (1-i)zoverline z\
(1+i)z^4 -(1-i)zoverline z &= 0\
z[(1+i)z^3-(1-i)overline z]&= 0
endaligned
$$

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

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

Hint

$$(1+i)z^4=(1-i)|z|^2iff z^4=dfrac1-i1+i|z|^2=-i|z|^2.$$

Taking modulus we have $|z|=0$ or $|z|=1.$ In the first case it is $z=0.$ In the second case $z=e^itheta.$ So, we have

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

begincasescos(4theta)=0\sin(4theta)=-1endcases

$$(1+i)z^4=(1-i)|z|^2iff iz^4=zbar ziffbegincasesz=0\iz^3=bar zendcases$$ $$iz^3=bar ziff i(a+ib)^3=(a-ib)iff ia^3 - 3 a^2 b -i 3 a b^2 + b^3=(a-ib)\iff b^3- 3 a^2 b +i(a^3 - 3 a b^2)=a-ibiffbegincasesb^3- 3 a^2 b=a\a^3 - 3 a b^2=-bendcases$$

We can also notice that $|b^3- 3 a^2 b +i(a^3 - 3 a b^2)|=sqrt(a^2+b^2)^3=sqrta^2+b^2=|a-ib|implies
|z|=begincases1\0endcases$, so it comes with terms with @mfl's answer