Find all four distinct fourth roots of -16i

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}

Learn how to find all four distinct fourth roots of -16i with detailed step-by-step instructions. This problem covers complex numbers, polar form, and finding fourth roots of complex numbers. Suitable for advanced high school students.

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