Question

Follow steps (a), (b), and (c) above to find all the values of the indicate droots $\sqrt[\mathbf{6}]{\mathbf{-}\mathbf{1}}$.

Short answer

The value of $\sqrt[6]{-1}$is $\frac{\left(\pm \sqrt{3}\pm I\right)}{2}$ .

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

Step-by-step solution

Given Information

The given expression is$\sqrt[6]{-1}$

Definition of Complex Number

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

Z=a+ib

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

Find the value of r and  θ

The Complex number is in the form -1+0i.

x=-1,y=0

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

r=1

$\theta =\pi or3\pi ,5\pi ,7\pi ,9\pi ,11\pi ,13\pi ,....$

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

$$\begin{gathered} z^{\frac{1}{n}}={\left(r{e}^{i\theta }\right)}^{\frac{1}{n}} \\ =r^{\frac{1}{n}}{e}^{\frac{i\theta }{n}} \\ {z}^{\frac{1}{n}}=\sqrt[n]{r}\left(\cos \frac{\theta }{n}+i\sin \frac{\theta }{n}\right)\left(1\right) \end{gathered}$$

When n=6 , the equation becomes the 6th root of the complex number as:

$$\begin{gathered} z^{\frac{1}{6}}=r^{\frac{1}{6}}{e}^{\frac{i\theta }{6}} \\ r=1 \\ \theta =\frac{\pi }{6},\frac{3\pi }{6},\frac{5\pi }{6},\frac{7\pi }{6},\frac{9\pi }{6},\frac{11\pi }{6},\frac{13\pi }{6},....... \\ =\frac{\pi }{6},\frac{\pi }{2},\frac{5\pi }{6},\frac{7\pi }{6},\frac{3\pi }{6},\frac{11\pi }{6},\frac{13\pi }{6},....... \end{gathered}$$

Plotting the polar coordinate points on the graph

It is clear from the above graph that the points $\left(1,\frac{\pi }{6}\right)$and the point $\left(1,\frac{13\pi }{6}\right)$are the same.

The radius of the circle is 1 and equally spaced $\frac{\pi }{3}$apart.

$$\begin{gathered} r=1 \\ \theta =\frac{\pi }{2},\frac{2\pi }{3},\frac{\pi }{2},\frac{7\pi }{6},\frac{3\pi }{2},\frac{11\pi }{6} \end{gathered}$$

Hence, the value of $\sqrt[6]{-1}=\pm i,\frac{\pm \sqrt{3}\pm i}{2}$.