Chapter I, Problem 39E

To determine

To find: The roots of the cube root of i and sketch the roots in the complex plane.

Answer to Problem 39E

The roots of cube root of i are $w_k=1^{\frac{1}{3}}\left[\cos \left(\frac{\frac{\pi }{2}+2k\pi }{3}\right)+i\sin \left(\frac{\frac{\pi }{2}+2k\pi }{3}\right)\right]$ where $k=0,1,2$.

Explanation of Solution

Theorem used:

Roots of a complex number:

Let $z=r\left(\cos \theta +i\sin \theta \right)$ and n be a positive integer. Then $z$ has the $n$ distinct nth roots $w_k=r^{\frac{1}{n}}\left[\cos \left(\frac{\theta +2k\pi }{n}\right)+i\sin \left(\frac{\theta +2k\pi }{n}\right)\right]$ where $k=0,1,2,\dots ,n-1$.

Calculation:

Rewrite the complex number i in polar form.

The polar form of the complex number $z=a+bi$ is $z=r\left(\cos \theta +i\sin \theta \right)$ where $r=\left|z\right|=\sqrt{a^2+b^2}$  and $\tan \theta =\frac{b}{a}$.

Consider the complex number i. Here, $i=0+i$. So, $a=0$, $b=1$ and $n=3$.

Obtain the argument of the complex number i.

$\begin{array}{l}\tan \theta =\frac{1}{0}\\ \theta ={\tan }^{-1}\left(\frac{1}{0}\right)\\ \theta =\frac{\pi }{2}\end{array}$

Thus, the argument of the complex number i is $\theta =\frac{\pi }{2}$.

Obtain the modulus of the complex number i.

$\begin{array}{c}r=\left|i\right|\\ =\sqrt{0^2+{\left(1\right)}^2}\\ =\sqrt{1}\\ =1\end{array}$

Thus, the value of $r=1$.

Therefore, the polar form of the complex number i is $1=\left(\cos \frac{\pi }{2}+i\sin \frac{\pi }{2}\right)$.

By the above theorem, the roots of cube root of i are $w_k=1^{\frac{1}{3}}\left[\cos \left(\frac{\frac{\pi }{2}+2k\pi }{3}\right)+i\sin \left(\frac{\frac{\pi }{2}+2k\pi }{3}\right)\right]$ where $k=0,1,2$.

Use online calculator to sketch the roots in the complex plane as shown below in Figure 1.

From figure 1, it is observed that all cube root of i form a triangle on complex plane.