Question

In Problems 53–60, find all the complex roots. Leave your answers in polar form with the argument in degrees.

58. The complex cube roots of$-8$.

Short answer

The complex cube roots of $-8$are$2(\cos 60+i\sin 60),2(\cos 180+i\sin 180),2(\cos 300+i\sin 300)$.

Step-by-step solution

Step 1. Rewrite -8 in the polar form.

The magnitude of $-8$is

$\sqrt{{(-8)}^2+0}=8$

which gives

$$\begin{gathered} -8=8(-1-0i) \\ =8(\cos 180°+i\sin 180°) \end{gathered}$$

Step 2. Using the Theorem of complex roots.

We get

$$\begin{gathered} z_k=\sqrt[3]{8}\left[\cos \right(\frac{180}{3}+\frac{360k}{3})+i\sin (\frac{180}{3}+\frac{360k}{3});k=0,1,2 \\ =2[\cos (60+120k)+i\sin (60+120k)] \end{gathered}$$

Step 3. Substitute k=0,1,2 in zk=2(cos(60+120k)+isin(60+120k)).

We get

$$\begin{gathered} z_0=2(\cos 60+i\sin 60) \\ {z}_1=2\left(\cos \right(60+120)+i\sin (60+120\left)\right) \\ =2(\cos 180+i\sin 180) \\ {z}_2=2\left(\cos \right(60+240)+i\sin (60+240\left)\right) \\ =2(\cos 300+i\sin 300) \end{gathered}$$

So we have found the complex cube roots of$-8$.