Answer
The complex cube roots of the given complex numbers in rectangular form are,
${{z}_{0}}=\frac{\sqrt{3}}{2}+\frac{1}{2}i,\ {{z}_{1}}=-\frac{\sqrt{3}}{2}+\frac{1}{2}i,\ {{z}_{2}}=0-1i$.
Work Step by Step
Consider the given complex number to write in the polar form,
$z=0+i$ ...... (1)
For a complex number $z=x+iy$, the polar form is given by,
$z=r\left( \cos \theta +i\sin \theta \right)$ ...... (2)
Here, $x=r\cos \theta \ \text{ and }\ y=r\sin \theta $, dividing the value of y by x, to get,
$\tan \theta =\frac{y}{x}$ ...... (3)
Also, the value of r is called the moduli of the complex number, given by,
$r=\sqrt{{{x}^{2}}+{{y}^{2}}}=\sqrt{{{\left( 0 \right)}^{2}}+{{\left( 1 \right)}^{2}}}=\sqrt{0+1}=1$
For any complex number $z=x+iy$, the sign of the value of x and y determine in which quadrant the value of $z=x+iy$ would lie.
If the value of x is on the positive side and the value of y is positive, then the angle $\theta $ lies in the first quadrant having the value of $\theta $ as $\theta $.
If the value of x lies on the negative side and the value of y is positive, then the angle $\theta $ lies in the second quadrant having the value of $\theta $ as $\pi -\theta $.
If the value of x lies on the negative side and the value of y is positive, then the angle $\theta $ lies in the second quadrant having the value of $\theta $ as $\pi +\theta $.
If the value of x is on the positive side and the value of y is negative, then the angle $\theta $ lies in the fourth quadrant having the value of $\theta $ as $2\pi -\theta $.
For the given complex number, using (1), (3) to get,
$\tan \theta =\frac{y}{x}=\frac{1}{0}=\text{Undefined}$
For the given complex number,
$\theta =\frac{\pi }{2}$
Using (2), to get,
$z=\left( \cos \frac{\pi }{2}+i\sin \frac{\pi }{2} \right)$
The polar form of the complex number is,
$z=\left( \cos \frac{\pi }{2}+i\sin \frac{\pi }{2} \right)$
Consider the given complex number,
$z=\left( \cos \frac{\pi }{2}+i\sin \frac{\pi }{2} \right)$
For any complex number of the type,
$z=r\cos \theta +i\sin \theta $
If n is a positive integer, then ${{z}^{n}}$ is,
${{z}_{k}}=\sqrt[n]{r}\left[ \cos \left( \frac{\theta +2\pi k}{n} \right)+i\sin \left( \frac{\theta +2\pi k}{n} \right) \right]$ ...... (4)
Now, for the given complex number, using (4) to get,
$\begin{align}
& {{z}_{k}}=\sqrt[n]{r}\left[ \cos \left( \frac{\theta +2\pi k}{n} \right)+i\sin \left( \frac{\theta +2\pi k}{n} \right) \right] \\
& {{z}_{0}}=\sqrt[3]{1}\left[ \cos \frac{\frac{\pi }{2}+2\pi \times 0}{3}+i\sin \frac{\frac{\pi }{2}+2\pi \times 0}{3} \right] \\
& {{z}_{0}}=1\left[ \cos \frac{\pi }{6}+i\sin \frac{\pi }{6} \right] \\
& {{z}_{0}}=\frac{\sqrt{3}}{2}+\frac{1}{2}i \\
\end{align}$
Now, for the other roots, substituting the value of $k=0,1,2$
That is.,
$\begin{align}
& {{z}_{k}}=\sqrt[n]{r}\left[ \cos \left( \frac{\theta +2\pi k}{n} \right)+i\sin \left( \frac{\theta +2\pi k}{n} \right) \right] \\
& {{z}_{1}}=\sqrt[3]{1}\left[ \cos \frac{\frac{\pi }{2}+2\pi \times 1}{3}+i\sin \frac{\frac{\pi }{2}+2\pi \times 1}{3} \right] \\
& {{z}_{1}}=1\left[ \cos \frac{5\pi }{3}+i\sin \frac{5\pi }{3} \right] \\
& {{z}_{1}}=-\frac{\sqrt{3}}{2}+\frac{1}{2}i \\
\end{align}$
Similarly,
$\begin{align}
& {{z}_{k}}=\sqrt[n]{r}\left[ \cos \left( \frac{\theta +2\pi k}{n} \right)+i\sin \left( \frac{\theta +2\pi k}{n} \right) \right] \\
& {{z}_{2}}=\sqrt[3]{1}\left[ \cos \frac{\frac{\pi }{2}+2\pi \times 2}{3}+i\sin \frac{\frac{\pi }{2}+2\pi \times 2}{3} \right]\ \ \ \\
& {{z}_{2}}=1\left[ \cos \frac{9\pi }{3}+i\sin \frac{9\pi }{3} \right] \\
& {{z}_{2}}=-i \\
\end{align}$
Therefore, the cube roots of the complex numbers in the polar form are,
${{z}_{0}}=\frac{\sqrt{3}}{2}+\frac{1}{2}i,\ {{z}_{1}}=-\frac{\sqrt{3}}{2}+\frac{1}{2}i,\ {{z}_{2}}=0-1i$
