Question

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

57. The complex fourth roots of$-16i$.

Short answer

The complex fourth roots of$-16i$are$2(\cos 67.5+i\sin 67.5),2(\cos 157.5+i\sin 157.5),2(\cos 247.5+i\sin 247.5),2(\cos 337.5+i\sin 337.5)$.

Step-by-step solution

Step 1. Rewrite -16i in the polar form.

The magnitude of $-16i$is

$$\begin{gathered} \sqrt{0^2+{(-16)}^2}=\sqrt{256} \\ =16 \end{gathered}$$

which gives

$$\begin{gathered} -16i=16(0-i) \\ =16(\cos 270°-i\sin 270°) \end{gathered}$$

Step 2. Using the theorem of complex roots.

We get

$$\begin{gathered} z_k=\sqrt[4]{16}\left[\cos \right(\frac{270}{4}+\frac{360k}{4})+i\sin (\frac{270}{4}+\frac{360k}{4}\left)\right];k=0,1,2,3 \\ =2\left[\cos \right(67.5°+90°k)+i\sin (67.5°+90°k\left)\right] \end{gathered}$$

Step 3. Substitute k=0,1,2,3 in zk=2[cos(67.5+90k)+isin(67.5+90k)].

We get

$$\begin{gathered} z_0=2(\cos 37.5+i\sin 67.5) \\ {z}_1=2\left[\cos \right(67.5+90)+i\sin (67.5+90\left)\right] \\ =2(\cos 157.5+i\sin 157.5) \\ {z}_2=2\left[\cos \right(67.5+180)+i\sin (67.5+180\left)\right] \\ =2(\cos 247.5+i\sin 247.5) \\ {z}_3=2\left[\cos \right(67.5+270)+i\sin (67.5+270\left)\right] \\ =2(\cos 337.5+i\sin 337.5) \end{gathered}$$

So we have found the complex fourth roots of $-16i$.