To find all four distinct fourth roots of $-16i$, we'll proceed as follows:

1. Express the complex number in polar form

First, we express $-16i$ in polar form. The polar form of a complex number $z = x + yi$ is given by:

$z = r \left(\cos \theta + i \sin \theta \right)$

where:

• $r = \sqrt{x^2 + y^2}$ is the modulus of the complex number.
• $\theta = \tan^{-1}\left(\frac{y}{x}\right)$ is the argument (angle) of the complex number.

For $-16i$, we have:

• $x = 0$ and $y = -16$.

So, the modulus $r$ is:

$r = \sqrt{0^2 + (-16)^2} = \sqrt{256} = 16$

Next, we calculate the argument $\theta$. Since the point is purely imaginary and lies on the negative imaginary axis:

$\theta = \tan^{-1}\left(\frac{-16}{0}\right) = -\frac{\pi}{2}$

Therefore, the polar form of $-16i$ is:

$-16i = 16 \left(\cos\left(-\frac{\pi}{2}\right) + i \sin\left(-\frac{\pi}{2}\right)\right)$

2. Find the fourth roots

The fourth roots of a complex number in polar form $r \left(\cos \theta + i \sin \theta \right)$ are given by:

$z_k = \sqrt[4]{r} \left( \cos\left(\frac{\theta + 2k\pi}{4}\right) + i \sin\left(\frac{\theta + 2k\pi}{4}\right) \right) \quad \text{for} \quad k = 0, 1, 2, 3$

For our case:

• $r = 16$, so $\sqrt[4]{r} = \sqrt[4]{16} = 2$.
• $\theta = -\frac{\pi}{2}$.

The roots are then:

$z_k = 2 \left( \cos\left(\frac{-\frac{\pi}{2} + 2k\pi}{4}\right) + i \sin\left(\frac{-\frac{\pi}{2} + 2k\pi}{4}\right) \right)$

3. Calculate each root

For $k = 0$:

[ z_0 = 2 \left( \cos\left(\frac{-\frac{\pi}{2} + 0}{4}\right) + i \sin\left(\frac{-\frac{\pi}{2}