Q16P - Mathematical Methods in Physical Sciences

Question

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

Step-by-Step Solution

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 1: Given Information

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

Step 2: 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 $\sqrt{-1}$.

Step 3: 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

$\theta =\pi or 3\pi ,5\pi ,7\pi ,9\pi ,11\pi ,13\pi ,....$

 

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:

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

Step 4: 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 $\sqrt[6]{-1}=\pm i,\frac{\pm \sqrt{3}\pm i}{2}$.