Chapter 9.3, Problem 60AYU

To determine

The complex fifth roots of $-i$ in exponential form.

Answer to Problem 60AYU

Solution:

The complex fifth roots of $-i$ are $z_0=e^{i\cdot \left(\frac{3\pi }{10}\right)},z_1=e^{i\cdot \left(\frac{7\pi }{10}\right)},z_2=e^{i\cdot \left(\frac{11\pi }{10}\right)},z_3=e^{i\cdot \left(\frac{15\pi }{10}\right)}$ and $z_4=e^{i\cdot \left(\frac{19\pi }{10}\right)}$.

Explanation of Solution

Given information:

The complex number, $-i$

Explanation:

To express $-i$ in exponential form $z=r{e}^{\theta }$ using degrees, find $r$ and the argument $\theta$.

The point $z=-i$ is located on negative $y-$ axis.

Comparing $z=-i$ with $z=x+yi$,

$\Rightarrow x=0$ and $y=-1$

Substitute the values of $x=0$ and $y=-1$ in the formula $r=\sqrt{x^2+y^2}$

$r=\sqrt{{\left(0\right)}^2+{\left(-1\right)}^2}=\sqrt{1}=1$

And $0\le \theta \le 2\pi$ is argument of $z$, such that $\sin \theta =\frac{y}{r}$ and $\cos \theta =\frac{x}{r}$

$\Rightarrow \sin \theta =\frac{-1}{1}=-1$ and $\cos \theta =0$

$\Rightarrow \tan \theta =\frac{-1}{0}$

$\Rightarrow \theta =\frac{3\pi }{2}$

Therefore, the polar form of a complex number $z=-i$ is $z=1e^{i\frac{3\pi }{2}}$.

By complex root theorem,

Let $w=r{e}^{i\theta }$ is a complex number and $n\ge 2$ be an integer. If $w\ne 0$, then there are $n$ distinct complex nth root of $w$, given by the formula

$z_k=\sqrt[n]{r}\left[e^{i\left(\frac{1}{n}\right)\left(\theta +2\pi k\right)}\right]$, where $k=0,1,2,\mathrm{...},n-1$.

To find the complex fifth roots of $z=1e^{i\frac{3\pi }{2}}$ by using the above formula,

Since, $n=5\ge 2$ and $z=1e^{i\frac{3\pi }{2}}\ne 0$

The fifth roots of $z=1e^{i\frac{3\pi }{2}}$ are as below:

$z_k=z_k=\sqrt[5]{1}\left[e^{i\cdot \frac{1}{5}\cdot \left(\frac{3\pi }{2}+2\pi k\right)}\right]$, where $k=0,1,2,3,4$

$z_k=\sqrt[5]{1}\left[e^{i\cdot \left(\frac{3\pi }{10}+\frac{4\pi k}{10}\right)}\right]$ where $k=0,1,2,3,4$

For $k=0$

$z_0=\sqrt[5]{1}\left[e^{i\cdot \left(\frac{3\pi }{10}+\frac{4\pi \left(0\right)}{10}\right)}\right]=\sqrt[5]{1}\cdot e^{i\cdot \left(\frac{3\pi }{10}\right)}=e^{i\cdot \left(\frac{3\pi }{10}\right)}$

For $k=1$

$z_1=\sqrt[5]{1}\left[e^{i\cdot \left(\frac{3\pi }{10}+\frac{4\pi }{10}\right)}\right]=\sqrt[5]{1}\cdot e^{i\cdot \left(\frac{7\pi }{10}\right)}=e^{i\cdot \left(\frac{7\pi }{10}\right)}$

For $k=2$

$\Rightarrow z_2=\sqrt[5]{1}\left[e^{i\cdot \left(\frac{3\pi }{10}+\frac{4\cdot 2\pi }{10}\right)}\right]=\sqrt[5]{1}\cdot e^{i\cdot \left(\frac{11\pi }{10}\right)}=e^{i\cdot \left(\frac{11\pi }{10}\right)}$

For $k=3$

$\Rightarrow z_3=\sqrt[5]{1}\left[e^{i\cdot \left(\frac{3\pi }{10}+\frac{4\cdot 3\pi }{10}\right)}\right]=\sqrt[5]{1}\left[e^{i\cdot \left(\frac{3\pi }{10}+\frac{12\pi }{10}\right)}\right]=\sqrt[5]{1}\cdot e^{i\cdot \left(\frac{15\pi }{10}\right)}=e^{i\cdot \left(\frac{15\pi }{10}\right)}$

For $k=4$

$\Rightarrow z_4=\sqrt[5]{1}\left[e^{i\cdot \left(\frac{3\pi }{10}+\frac{4\cdot 4\pi }{10}\right)}\right]=\sqrt[5]{1}\left[e^{i\cdot \left(\frac{3\pi }{10}+\frac{16\pi }{10}\right)}\right]=\sqrt[5]{1}\cdot e^{i\cdot \left(\frac{19\pi }{10}\right)}=e^{i\cdot \left(\frac{19\pi }{10}\right)}$

Therefore, the complex fifth roots of $-i$ are

$z_0=e^{i\cdot \left(\frac{3\pi }{10}\right)},z_1=e^{i\cdot \left(\frac{7\pi }{10}\right)},z_2=e^{i\cdot \left(\frac{11\pi }{10}\right)},z_3=e^{i\cdot \left(\frac{15\pi }{10}\right)}$ and $z_4=e^{i\cdot \left(\frac{19\pi }{10}\right)}$.