Answer
$2cis45^\circ (\sqrt{2} + \sqrt{2}i)$,
$2cis135^\circ (- \sqrt{2} + \sqrt{2}i)$,
$2cis225^\circ (- \sqrt{2} - \sqrt{2}i)$ and
$2cis315^\circ (\sqrt{2} - \sqrt{2}i)$ are the four fourth roots of $-16$.
Work Step by Step
$-16$ is at $180^\circ$ with absolute value $16$.
The equivalent trigonometric form of $-16$ is $16cis180^\circ$.
Suppose the fourth root of $-16$ is represented by $rcis\alpha$,
$(rcis\alpha)^4$ = $-16$ = $16cis180^\circ$
By De Moivre’s theorem, this equation becomes $r^4cis4\alpha = 16cis180^\circ$
By equating, $r^4 = 16$, $r = 2$, and $cos4\alpha = cos180^\circ, sin4\alpha = sin180^\circ$
For these equations to be satisfied, $4\alpha$ must represent an angle that is coterminal with $180^\circ$.
Therefore, we must have $\alpha = \frac{180^\circ + 360^\circ\cdot k}{4}$ and k takes on the integer values 0, 1, 2 and 3.
When $k = 0, \alpha = 45^\circ$,
$k = 1, \alpha = 135^\circ$,
$k = 2, \alpha = 225^\circ$,
$k = 3, \alpha = 315^\circ$
Hence, when $k = 0,$ the root is $2cis45^\circ$,
$k = 1,$ the root is $2cis135^\circ$,
$k = 2,$ the root is $2cis225^\circ$,
$k = 3,$ the root is $2cis315^\circ$
Thus,
$2cis45^\circ (\sqrt{2} + \sqrt{2}i)$,
$2cis135^\circ (- \sqrt{2} + \sqrt{2}i)$,
$2cis225^\circ (- \sqrt{2} - \sqrt{2}i)$ and
$2cis315^\circ (\sqrt{2} - \sqrt{2}i)$ are the four fourth roots of $-16$.
