Chapter 9.3, Problem 57AYU

In Problems 55-60, find all the complex roots. Leave your answers in polar form with the argument in degrees.

The complex cube roots of $-16i$

To determine

To find: Complex fourth roots. Leave your answers in polar form with the argument in degrees.

Answer to Problem 57AYU

Solution:

$z_0=2\left(\cos \left(67.5°\right)+i\sin \left(67.5°\right)\right)$, $z_1=2\left(\cos \left(157.5°\right)+i\sin \left(157.5°\right)\right)$

$z_2=\sqrt[4]{2}\left(\cos \left(247.5°\right)+i\sin \left(247.5°\right)\right)$, $z_3=\sqrt[4]{2}\left(\cos \left(337.5°\right)+i\sin \left(337.5°\right)\right)$

Explanation of Solution

Given:

$-16i$

Calculation:

We can use here De Moivre's theorem, which states that if $z=r\left(\cos \theta +i\sin \theta \right)$, then $z^n=r^n\left(\cos n\theta +i\sin n\theta \right)$.

It may be worth mentioning that the theorem is valid for fractions as well. As we are going to find cube roots, what we seek is ${\left(-16i\right)}^{\frac{1}{4}}$.

For that let us first write $-16i$ in polar form.

$r=\sqrt{{\left(0\right)}^2+{\left(-16\right)}^2}=\sqrt{256}=16$ and $\theta ={\tan }^{-1}\left(\frac{y}{x}\right)={\tan }^{-1}\left(0\right)=270°$

$-16i=16\left(\cos \left(270°\right)+i\sin \left(270°\right)\right)$

${\left(-16i\right)}^{\frac{1}{4}}={\left(16\left(\cos \left(270°\right)+i\sin \left(270°\right)\right)\right)}^{\frac{1}{4}}$

$={16}^{\frac{1}{4}}\left(\cos \left(\frac{360°k+270°}{4}\right)+i\sin \left(\frac{360°k+270°}{4}\right)\right)k=0,1,2,3$

$=2\left(\cos \left(\frac{360°k+270°}{4}\right)+i\sin \left(\frac{360°k+270°}{4}\right)\right)$

So, $z_0=2\left(\cos \left(\frac{0+270°}{4}\right)+i\sin \left(\frac{0+270°}{4}\right)\right)=2\left(\cos \left(67.5°\right)+i\sin \left(67.5°\right)\right)$.

$z_1=2\left(\cos \left(\frac{360°+270°}{4}\right)+i\sin \left(\frac{360°+270°}{4}\right)\right)=2\left(\cos \left(157.5°\right)+i\sin \left(157.5°\right)\right)$

$z_2=2\left(\cos \left(\frac{720°+270°}{4}\right)+i\sin \left(\frac{720°+270°}{4}\right)\right)=\sqrt[4]{2}\left(\cos \left(247.5°\right)+i\sin \left(247.5°\right)\right)$

$z_3=\sqrt[4]{2}\left(\cos \left(\frac{1080°+270°}{4}\right)+i\sin \left(\frac{1080°+270°}{4}\right)\right)=\sqrt[4]{2}\left(\cos \left(337.5°\right)+i\sin \left(337.5°\right)\right)$