Chapter 14, Problem 15CLR

To determine

To find: The cube roots of $-8i$ in $x+yi$ form.

Answer to Problem 15CLR

The required roots in $x+yi$ form $2i$ , $\sqrt{3}-i$ , and $-\sqrt{3}-i$ .

Explanation of Solution

Given:

The expression is $-8i$ .

Calculation:

Let $z=0-8i$ .

Compare the expression $z=0-8i$ with $z=x+yi$ .

  $\begin{array}{c}x=0\\ y=-8\end{array}$

Substitute 0 for $x$ and $-8$ for $y$ in the equation $r=\sqrt{x^2+y^2}$ to find $r$ .

  $\begin{array}{c}r=\sqrt{{\left(0\right)}^2+{\left(-8\right)}^2}\\ =\sqrt{0+64}\\ =\sqrt{64}\\ =8\end{array}$

Find the angle by using the formula $\theta =\arctan \left(\frac{y}{x}\right)$ .

  $\begin{array}{c}\theta =\arctan \left(\frac{-8}{0}\right)\\ =90°\end{array}$

So, the polar form of $z=0-8i$ is $8\left(\cos 90°-i\sin 90°\right)$ .

Consider $z=r\left(\cos \theta +i\sin \theta \right)$ as a root of $-1$ .

The DeMoivre’s theorem states that if $z^n={\left[r\left(\cos \theta +i\sin \theta \right)\right]}^n$ and $n$ is a positive integer then $z^n=r^n\left(\cos n\theta +i\sin n\theta \right)$ . So,

  $z^3=r^2\left(\cos 3\theta +i\sin 3\theta \right)$

Since $z^3$ is $-8i$ , $r^3\left(\cos 3\theta +i\sin 3\theta \right)=8\left(\cos 90°-i\sin 90°\right)$ .

On computing, get $r^3$ as $8$ , $\cos 3\theta$ as $90°$ , and $\sin 3\theta$ as $90°$ and $r=0$ .

Thus $3\theta$ differs from $90°$ by an integral multiple of $360°$ .

  $\begin{array}{c}3\theta =90°+k\cdot 360°\\ \theta =30°+k\cdot 120°\end{array}$

Substitute 0, 1 and $2$ for $k$ and evaluate the following.

  $\begin{array}{c}\theta =30°+0\cdot 120°\\ =30°\\ \theta =30°+1\cdot 120°\\ =150°\end{array}$

And,

  $\begin{array}{c}\theta =30°+2\cdot 120°\\ =270°\end{array}$

The three cube roots in polar form are $2\left(\cos 30°-i\sin 30°\right)$ , $2\left(\cos 150°-i\sin 150°\right)$ , and $2\left(\cos 270°-i\sin 270°\right)$ .

Find the root is $x+yi$ form as follows.

  $\begin{array}{c}2\left(\cos 30°-i\sin 30°\right)=\sqrt{3}-i\\ 2\left(\cos 150°-i\sin 150°\right)=-\sqrt{3}-i\\ 2\left(\cos 270°-i\sin 270°\right)=2i\end{array}$

Therefore, the required roots in $x+yi$ form $2i$ , $\sqrt{3}-i$ , and $-\sqrt{3}-i$ .