Evaluate $z^4-(1-i)^2=0$

Question

Will it be correct to look at it as

$z^4=(1-i)^2$

so

$r^4e^{4\theta i}=2e^{-\frac{\pi}{2}i}$

So

$$re^{i\theta}=\sqrt[4]{2}e^{\frac{-\pi i/2+2 \pi i k}{4}}$$ where $k=0,1,2,3$

Answer 1

Hint:

Using $A^2-B^2=(A+B)(A-B)$ is is equivalent to:

$$ z^2=1-i \quad \mbox{or}\quad z^2=i-1 $$

Answer 2

$$z^4 = (1-i)^2=(\sqrt{2}\exp\left(\frac{-i\pi}{4} \right))^2=2\exp\left(\frac{-i\pi}{2} \right)=2\exp\left(i\left(\frac{-\pi}{2}+2k\pi \right)\right)$$

$$z=2^\frac14\exp\left(i\left(\frac{-\pi}{8}+\frac{k\pi}2 \right)\right)$$

where $k=0,1,2,3$.