Chapter 9.3, Problem 58AYU

To determine

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

Answer to Problem 58AYU

Solution:

$z_0=2\left(\cos \left(60°\right)+i\sin \left(60°\right)\right)$, $z_1=2\left(\cos \left(180°\right)+i\sin \left(180°\right)\right)$

$z_2=2\left(\cos \left(300°\right)+i\sin \left(300°\right)\right)$

Explanation of Solution

Given:

$-8$

Calculation:

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

$r=\sqrt{{\left(-8\right)}^2+0}=\sqrt{64}=8$ and $\theta ={\tan }^{-1}\left(\frac{y}{x}\right)={\tan }^{-1}\left(\frac{-8}{0}\right)={\tan }^{-1}\left(\infty \right)=180°$

$-8=8\left(\cos \left(180°\right)+i\sin \left(180°\right)\right)$

${\left(-8\right)}^{\frac{1}{3}}={\left(8\left(\cos \left(180°\right)+i\sin \left(180°\right)\right)\right)}^{\frac{1}{3}}$

$=2\left(\cos \left(\frac{360k+180°}{3}\right)+i\sin \left(\frac{360k+180°}{3}\right)\right)k=0,1,2$

So, $z_0=2\left(\cos \left(\frac{0+180°}{3}\right)+i\sin \left(\frac{0+180°}{3}\right)\right)=2\left(\cos \left(60°\right)+i\sin \left(60°\right)\right)$.

$z_1=2\left(\cos \left(\frac{360°+180°}{3}\right)+i\sin \left(\frac{360°+180°}{3}\right)\right)=2\left(\cos \left(180°\right)+i\sin \left(180°\right)\right)$

$z_2=2\left(\cos \left(\frac{720°+180°}{3}\right)+i\sin \left(\frac{720°+180°}{3}\right)\right)=2\left(\cos \left(300°\right)+i\sin \left(300°\right)\right)$