Question

The roots of the equation $z^{4}+1=0$ are

• A. $(\pm 1 \pm i)$
• B. $(\pm 2 \pm 2 i)$
• C. $\frac{1}{\sqrt{2}}(\pm 1 \pm i)$
• D. None of these

Answer 1

Answer: (C)

We have, $z^{4}+1=0 \Rightarrow z^{4}=-1$
$\Rightarrow z=(\cos \pi+i \sin \pi)^{1 / 4} $
$\Rightarrow z=\cos \frac{1}{4}(2 k \pi+\pi)+i \sin \frac{1}{4}(2 k \pi+\pi), k=0,1,2,3 $
$\Rightarrow z=\cos \frac{\pi}{4}+i \sin \frac{\pi}{4}, \cos \frac{3 \pi}{4}+i \sin \frac{3 \pi}{4} $
$\cos \frac{5 \pi}{4}+i \sin \frac{5 \pi}{4}, \cos \frac{7 \pi}{4}+i \sin \frac{7 \pi}{4}$
$=\frac{1}{\sqrt{2}}(1+i), \frac{1}{\sqrt{2}}(-1+i), \frac{1}{\sqrt{2}}(-1-i), \frac{1}{\sqrt{2}}(1-i)$
Hence, the four roots of $z^{4}+1=0$ are $\frac{1}{\sqrt{2}}(\pm 1 \pm i)$.