Answer to Question #350645 in Complex Analysis for John

Question

Find the four roots of z⁴=-1

Answer 1

$z^4=-1$

$z=\sqrt[4]{-1}$

The roots of a complex number are also given by a formula $\sqrt[n]{z}=\sqrt[n]{|z|}\left(\cos\frac{\varphi+2\pi k}{n}+i\sin\frac{\varphi+2\pi k}{n}\right)$, where $k=\overline{0,n-1}$.

$|-1|=1$, $\varphi=\pi$

$k=0$:

$z_1=\sqrt[4]{-1}=\sqrt[4]{1}\left(\cos\frac{\pi+2\pi\cdot0}{4}+i\sin\frac{\pi+2\pi\cdot0}{4}\right)=1\cdot\left(\cos\frac{\pi}{4}+i\sin\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2}$

$k=1$:

$z_2=\sqrt[4]{-1}=\sqrt[4]{1}\left(\cos\frac{\pi+2\pi\cdot1}{4}+i\sin\frac{\pi+2\pi\cdot1}{4}\right)=1\cdot\left(\cos\frac{3\pi}{4}+i\sin\frac{3\pi}{4}\right)=\frac{\sqrt{2}}{2}-i\frac{\sqrt{2}}{2}$

$k=2$:

$z_3=\sqrt[4]{-1}=\sqrt[4]{1}\left(\cos\frac{\pi+2\pi\cdot2}{4}+i\sin\frac{\pi+2\pi\cdot2}{4}\right)=1\cdot\left(\cos\frac{5\pi}{4}+i\sin\frac{5\pi}{4}\right)=-\frac{\sqrt{2}}{2}-i\frac{\sqrt{2}}{2}$

$k=3$:

$z_4=\sqrt[4]{-1}=\sqrt[4]{1}\left(\cos\frac{\pi+2\pi\cdot3}{4}+i\sin\frac{\pi+2\pi\cdot3}{4}\right)=1\cdot\left(\cos\frac{7\pi}{4}+i\sin\frac{7\pi}{4}\right)=-\frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2}$

So $z_1=\frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2}$, $z_2=\frac{\sqrt{2}}{2}-i\frac{\sqrt{2}}{2}$, $z_3=-\frac{\sqrt{2}}{2}-i\frac{\sqrt{2}}{2}$, $z_4=-\frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2}$