First, we must write \(-16 i\) in trigonometric form.
The modulus is \(16\), and the argument is \(\frac{3\pi}{2}\). So the trigonometric form of \(-16 i\) is \(16 \left( \cos\frac{3\pi}{2} + i \sin\frac{3\pi}{2}\right) \).
There are always four fourth roots.
The modulus of each of the fourth roots will be the fourth root of the modulus of \(-16 i\). So the modulus of each of the fourth roots will be \(\sqrt[4]{16}=2\).
The argument of the first of the four fourth roots will be one-fourth of the angle for \(-16 i\). The complex number \(-16 i\) lies on the negative imaginary axis (the negative \(y\)-axis) therefore it lies on the terminal side of the angle \(\frac{3\pi}{2}\). So the angle of the first of the four fourth roots will be one-fourth of \(\frac{3\pi}{2}\)--that is, \(\frac{3\pi}{8}\).
The four fourth roots will lie evenly spaced around the unit circle, so they will lie an angular distance of \(\frac{\pi}{2}\) radians apart.
Therefore, the four arguments associated with the four fourth roots will be \(\frac{3\pi}{8}\), \(\frac{7\pi}{8}\), \(\frac{11\pi}{8}\), and \(\frac{15\pi}{8}\).
So the four fourth roots of \(-16 i\) are:
\(w_0=2\left( \cos\frac{3\pi}{8}+i\sin\frac{3\pi}{8}\right)\)
\(w_1=2\left( \cos\frac{7\pi}{8}+i\sin\frac{7\pi}{8}\right)\)
\(w_2=2\left( \cos\frac{11\pi}{8}+i\sin\frac{11\pi}{8}\right)\)
\(w_3=2\left( \cos\frac{15\pi}{8}+i\sin\frac{15\pi}{8}\right)\)