Question

Follow steps (a), (b), (c) above to find all the values of the indicated roots.

$\sqrt[\mathbf{3}]{\mathbf{-}\mathbf{8}\mathit{i}}$

Short answer

The value of $\sqrt[3]{-8i}$is $2i,\sqrt{3}-i$.

The graph used in this question to find the answer is shown below as:

Step-by-step solution

Given Information

The given expression is$\sqrt[3]{-8i}$.

Definition of Complex Number

Complex numbers comprise real numbers and imaginary numbers; a complex can be written in the form of:

$\mathit{z}\mathbf{=}\mathit{a}\mathbf{+}\mathit{i}\mathit{b}$

Here a and b are real numbers, and i is the imaginary number which is known as iota, whose value is $\sqrt{\mathbf{-}\mathbf{1}}$.

Find the value of r and θ

The Complex number is in the form $0-8i$.

$x=0,y=-8$

The polar coordinates of the point are in the form of $z=r{e}^{i\theta }$.

$\begin{array}{ll}r=8& \\ \theta =\frac{3\pi }{2},0r\frac{7\pi }{2},\frac{11\pi }{2}& \frac{15\pi }{2},\dots \end{array}$

The equation $z=r{e}^{i\theta }$can also be written in another form.

role="math" localid="1658745524955" $$\begin{gathered} \begin{array}{l}{z}^{\frac{1}{n}}={\left(r{e}^{i\theta }\right)}^{\frac{1}{n}}\\ z^{\frac{1}{n}}=r^{\frac{1}{n}}{e}^{\frac{i\theta }{n}} \\ {z}^{\frac{1}{n}}=\sqrt[n]{r}\left(\cos \frac{\theta }{n}+i\sin \frac{\theta }{n}\right)...........\left(1\right)\end{array} \end{gathered}$$

When $n=3$, the equation becomes the$3^{rd}$ root of the complex number.

$$\begin{gathered} z^{\frac{1}{3}}=r^{\frac{1}{3}}{e}^{\frac{i\theta }{3}} \\ r=2 \\ \theta =\frac{3\pi }{6},\frac{7\pi }{6},\frac{11\pi }{6},\frac{15\pi }{6},... \\ =\frac{\pi }{2},\frac{7\pi }{6},\frac{11\pi }{6},\frac{5\pi }{2},... \end{gathered}$$

Plotting the polar coordinate points on the graph

It is clear from the above graph that the points $\left(2,\frac{\mathrm{\pi }}{2}\right)$and the point $\left(2,\frac{5\mathrm{\pi }}{2}\right)$are the same.

The radius of the circle is 2 and equally spaced $\frac{2\pi }{3}$apart.

$$\begin{gathered} r=1 \\ \theta =\frac{\pi }{2},\frac{7\pi }{6},\frac{11\pi }{6} \end{gathered}$$

Hence, the value of $\sqrt[3]{-8i}=2i,\pm \sqrt{3}-i$.