Q12P - Mathematical Methods in Physical Sciences

Question

Follow steps (a), (b), (c) above to find all the values of the indicated roots $\sqrt[\mathbf{3}]{\mathbf{-}\mathbf{1}}$.

Step-by-Step Solution

Answer

The value of   is $\sqrt[3]{-1} \mathrm{is} -1,\frac{1\pm \mathrm{i}\sqrt{3}}{2}$.

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

Step 1: Given Information

The given expression is $\sqrt[3]{-1}$.

Step 2: Definition of Complex Number

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

$\mathbf{z}\mathbf{=}\mathbf{a}\mathbf{+}\mathbf{ib}$

 

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

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 $z=r{e}^{i\theta }$ .

 

$$\begin{gathered} r=1 \\ \theta =\pi ,3\pi ,5\pi ,7\pi ,..... \end{gathered}$$

 

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

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

                                                                   

When n=3 , the equation becomes the root of the complex number as:

 $$\begin{gathered} Z^{\frac{1}{3}}=r^{\frac{1}{3}}{e}^{\frac{i\theta }{3}} \\ r=1 \\ \theta =\frac{\pi }{3},\frac{3\pi }{3},\frac{5\pi }{3},\frac{7\pi }{3},.... \\ =\frac{\pi }{3},\pi ,\frac{5\pi }{3},\frac{7\pi }{3},... \end{gathered}$$

Step 4: Plotting of the polar coordinate points on the graph

   

It is clear from the above graph that the points $\left(1,\frac{\pi }{3}\right)$ and the point $\left(1,\frac{7\pi }{3}\right)$are the same.

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

$$\begin{gathered} r=1 \\ \theta =\frac{\pi }{3},\pi ,\frac{5\pi }{3} \end{gathered}$$

 

Hence, the final answer is $\sqrt[3]{-1}=-1,\left(\frac{1\pm i\sqrt{3}}{2}\right)$.