Chapter 9.3, Problem 55AYU

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 $4-4\sqrt{3}i$

To determine

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

Answer to Problem 55AYU

Solution:

$z_0=\sqrt[4]{2}\left(\cos \left(75°\right)+i\sin \left(75°\right)\right)$, $z_1=\sqrt[4]{2}\left(\cos \left(165°\right)+i\sin \left(165°\right)\right)$

$z_2=\sqrt[4]{2}\left(\cos \left(255°\right)+i\sin \left(255°\right)\right)$, $z_3=\sqrt[4]{2}\left(\cos \left(345°\right)+i\sin \left(345°\right)\right)$

Explanation of Solution

Given:

$4-4\sqrt{3}i$

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 ${\left(4-4\sqrt{3}i\right)}^{\frac{1}{4}}$.

For that let us first write $4-4\sqrt{3}i$ in polar form.

$r=\sqrt{{\left(4\right)}^2+{(-4\sqrt{3})}^2}=\sqrt{64}=8$ and $\theta ={\tan }^{-1}\left(\frac{y}{x}\right)={\tan }^{-1}\left(\frac{-4\sqrt{3}}{4}\right)={\tan }^{-1}\left(-\sqrt{3}\right)=300°$

$4-4\sqrt{3}i=8\left(\cos \left(300°\right)+i\sin \left(300°\right)\right)$

${\left(4-4\sqrt{3}i\right)}^{\frac{1}{4}}={\left(8\left(\cos \left(300°\right)+i\sin \left(300°\right)\right)\right)}^{\frac{1}{4}}$

$=8^{\frac{1}{4}}\left(\cos \left(\frac{360°k+300°}{4}\right)+i\sin \left(\frac{360°k+300°}{4}\right)\right)k=0,1,2,3$

$=\sqrt[4]{8}\left(\cos \left(\frac{360°k+300°}{4}\right)+i\sin \left(\frac{360°k+300°}{4}\right)\right)$

So, $z_0=\sqrt[4]{2}\left(\cos \left(\frac{0+300°}{4}\right)+i\sin \left(\frac{0+300°}{4}\right)\right)=\sqrt[4]{2}\left(\cos \left(75°\right)+i\sin \left(75°\right)\right)$.

$z_1=\sqrt[4]{2}\left(\cos \left(\frac{360°+300°}{4}\right)+i\sin \left(\frac{360°+300°}{4}\right)\right)=\sqrt[4]{2}\left(\cos \left(165°\right)+i\sin \left(165°\right)\right)$

$z_2=\sqrt[4]{2}\left(\cos \left(\frac{720°+300°}{4}\right)+i\sin \left(\frac{720°+300°}{4}\right)\right)=\sqrt[4]{2}\left(\cos \left(255°\right)+i\sin \left(255°\right)\right)$

$z_3=\sqrt[4]{2}\left(\cos \left(\frac{1080°+300°}{4}\right)+i\sin \left(\frac{1080°+300°}{4}\right)\right)=\sqrt[4]{2}\left(\cos \left(345°\right)+i\sin \left(345°\right)\right)$