Question

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

56. The complex cube roots of$-8-8i$.

Short answer

The complex cube roots of$-8-8i$are$2\sqrt[6]{2}(\cos 75+i\sin 75),2\sqrt[6]{2}(\cos 195+i\sin 195),2\sqrt[6]{2}(\cos 315+i\sin 315)$.

Step-by-step solution

Step 1. Rewrite -8-8i in the polar form.

The magnitude of $-8-8i$is

$$\begin{gathered} \sqrt{{(-8)}^2+{(-8)}^2}=\sqrt{64+64} \\ =\sqrt{128} \\ =8\sqrt{2} \end{gathered}$$

which gives

role="math" localid="1646811578868" $$\begin{gathered} -8-8i=8\sqrt{2}(-\frac{1}{\sqrt{2}}-i\frac{1}{\sqrt{2}}) \\ =\sqrt{2}·2^3(\cos 225°-i\sin 225°) \end{gathered}$$

Step 2. Use the theorem of complex roots.

We get

$$\begin{gathered} z_k=\sqrt[3]{\sqrt{2}·2^3}\left[\cos \right(\frac{225}{3}+\frac{360k}{3})+i\sin (\frac{225}{3}+\frac{360k}{3}\left)\right];k=0,1,2 \\ =2\sqrt[6]{2}\left[\cos \right(75+120k)+i\sin (75+120k\left)\right] \end{gathered}$$

Step 3. Substitute k=0,1,2 in zk=226(cos(75+120k)+isin(75+120k)).

We get

$$\begin{gathered} z_0=2\sqrt[6]{2}(\cos 75+i\sin 75) \\ {z}_1=2\sqrt[6]{2}(\cos 75+120)+i\sin (75+120) \\ =2\sqrt[6]{2}(\cos 195+i\sin 195) \\ {z}_2=2\sqrt[6]{2}\left(\cos \right(75+240)+i\sin (75+240\left)\right) \\ =2\sqrt[6]{2}(\cos 315+i\sin 315) \end{gathered}$$

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