Solve $(2z-1)^5 - i = 0$

Question

Solve $(2z-1)^5 - i = 0$

I started by saying that $(2z-1)^5 = i$

$(2z-1) = \sqrt[5]i$

$z =$ $(\sqrt[5]i +1) \over 2$

$z^5 =$ $(i +1) \over 32$

$z^5 =$ $1 \over32$$ *(i +1)$

From there, it's quite simple.. showing $z^5$ as $r^5Cis(5o)$ with da-muaver and showing $(1+i)$ as $\sqrt2 $$ Cis({\phi\over 4})$

Answer 1

Hint :

$\cos \theta+i\sin \theta=\sqrt[5]{i}\Rightarrow (\cos \theta+i\sin \theta)^5=i\Rightarrow \cos 5\theta+i\sin5\theta=i\Rightarrow \theta = ??$

Answer 2

Notice that

$$i=\exp\left(\frac {i\pi}2\right)$$ so $$Z^5=i\iff Z=\exp\left(\frac {i\pi}{10}+\frac{2ik\pi}{5}\right),\quad k=0,\ldots,4$$ Now let $Z=2z-1$ and solve the given equation for $z$.