Q19P - Mathematical Methods in Physical Sciences

Question

Follow steps (a), (b), (c) above to find all the values of the indicate droots $\sqrt[\mathbf{3}]{\mathbf{i}}$.

Step-by-Step Solution

Answer

The value of  $\sqrt[3]{i} is -i,\frac{\left(\pm \sqrt{3}+i\right)}{2}.$

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

Step 1: Given Information

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

Step 2: Definition of Complex Number

Complex numbers have both 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 0+i ..

x=0,y=1

 

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

 

$$\begin{gathered} r=1 \\ \theta =\frac{\pi }{2},or\frac{5\pi }{2},\frac{9\pi }{2},\frac{13\pi }{2},... \end{gathered}$$

 

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 3rd the root of the complex number.

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

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 and equally spaced $\frac{2\pi }{3}$apart.

$$\begin{gathered} r=1 \\ \theta =\frac{\pi }{6},\frac{5\pi }{6},\frac{3\pi }{2} \end{gathered}$$

 

Hence,  the value of $\sqrt[3]{i}=-i,\frac{\sqrt{\left(\pm \sqrt{3} +i\right)}}{2}$.