Chapter 9.3, Problem 54AYU

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 $\sqrt{3}-i$

To determine

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

Answer to Problem 54AYU

Solution:

$z_0=\left(\cos \left(82.5°\right)+i\sin \left(82.5°\right)\right)$, $z_1=\sqrt[4]{2}\left(\cos \left(172.5°\right)+i\sin \left(172.5°\right)\right)$

$z_2=\sqrt[4]{2}\left(\cos \left(262.5°\right)+i\sin \left(262.5°\right)\right)$, $z_3=\sqrt[4]{2}\left(\cos \left(352.5°\right)+i\sin \left(352.5°\right)\right)$

Explanation of Solution

Given:

$\sqrt{3}-i$

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

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

$r=\sqrt{{\left(\sqrt{3}\right)}^2+{(-1)}^2}=\sqrt{4}=2$ and $\theta ={\tan }^{-1}\left(\frac{y}{x}\right)={\tan }^{-1}\left(\frac{-1}{\sqrt{3}}\right)=-30°=330°$

$\sqrt{3}-i=2\left(\cos \left(330°\right)+i\sin \left(330°\right)\right)$

${\left(\sqrt{3}-i\right)}^{\frac{1}{4}}={\left(2\left(\cos \left(330°\right)+i\sin \left(330°\right)\right)\right)}^{\frac{1}{4}}$

$=2^{\frac{1}{4}}\left(\cos \left(\frac{360°k+330°}{4}\right)+i\sin \left(\frac{360°k+330°}{4}\right)\right)k=0,1,2,3$

$=\sqrt[4]{2}\left(\cos \left(\frac{360°k+330°}{4}\right)+i\sin \left(\frac{360°k+330°}{4}\right)\right)$

So, $z_0=\sqrt[4]{2}\left(\cos \left(\frac{0+330°}{4}\right)+i\sin \left(\frac{0+330°}{4}\right)\right)=\sqrt[4]{2}\left(\cos \left(82.5°\right)+i\sin \left(82.5°\right)\right)$.

$z_1=\sqrt[4]{2}\left(\cos \left(\frac{360°+330°}{4}\right)+i\sin \left(\frac{360°+330°}{4}\right)\right)=\sqrt[4]{2}\left(\cos \left(172.5°\right)+i\sin \left(172.5°\right)\right)$

$z_2=\sqrt[4]{2}\left(\cos \left(\frac{720°+330°}{4}\right)+i\sin \left(\frac{720°+330°}{4}\right)\right)=\sqrt[4]{2}\left(\cos \left(262.5°\right)+i\sin \left(262.5°\right)\right)$

$z_3=\sqrt[4]{2}\left(\cos \left(\frac{1080°+330°}{4}\right)+i\sin \left(\frac{1080°+330°}{4}\right)\right)=\sqrt[4]{2}\left(\cos \left(352.5°\right)+i\sin \left(352.5°\right)\right)$