Complex Equation problem

Question

I have this problem to solve in $C$

$$(1-i)z^5-i=0$$

My solution

$$z^5=\frac{-1+i}{2} = \frac{-1}{2}+\frac{i}{2} $$

Now I need to find args.

$$|z|=r=\sqrt{a^2+b^2} \rightarrow \sqrt{\frac{1}{4}+\frac{1}{4}}=\frac{\sqrt{2}}{2}$$

$$tan(args)=\frac{b}{a} \rightarrow tan(args)= \frac{\frac{1}{2}}{\frac{-1}{2}}=\frac{2}{-2}=1$$

Therefore args=45 or 135 $\rightarrow$ 135.

For some reason I don't get the same $args$ and $|z|$ as the answer in the book.

Any ideas? Thanks

Answer 1

We have

$$z^5=\frac i{1-i}=\frac i2(1+i)=\frac1{\sqrt2}e^{i\frac\pi2} e^{i\frac\pi4}=\frac1{\sqrt2}e^{3i\frac\pi4}$$

Now notice that

$$z^n=R e^{i\theta}\iff z=\sqrt[n]R\exp\left(\frac{i\theta+2ik\pi}{n}\right),\quad k=0,\ldots,n-1$$

Answer 2

$$|z^5|=\frac 1{\sqrt{2}}=>|z|=2^{-\frac 52}$$ now you can use De Moivre's theorem to find the args.