Question

If $i{z}^4+1=0$, then $z$ can take the value

• A. $\frac{1+i}{\sqrt{2}}$
• B. $\cos \frac{\pi }{8}+I\sin \frac{\pi }{8}$
• C. $\frac{1}{4i}$
• D. $i$

Answer 1

Answer: (B)

$i{z}^4=-1$
$\Rightarrow z^4=-\frac{1}{i}$
$\Rightarrow z^4=i$
$\Rightarrow z=(i{)}^{1/4}=(0+i{)}^{1/4}$
$\Rightarrow z={\left(\cos \frac{\pi }{2}+i\sin \frac{\pi }{2}\right)}^{1/4}$
$\Rightarrow z=\cos \frac{\pi }{8}+i\sin \frac{\pi }{8}$
(using De Moirrels theorem)