Chapter 9, Problem 13CT

To determine

The complex cube roots of $-8+8\sqrt{3}i$ in exponential form. Then plot them in complex plane.

Answer to Problem 13CT

Solution:

The complex cube roots of $-8+8\sqrt{3}i$ are $z_0=2\sqrt[3]{2}\cdot e^{i\cdot \left(\frac{2\pi }{9}\right)}$ , $z_1=2\sqrt[3]{2}\cdot e^{i\cdot \left(\frac{8\pi }{9}\right)}$ , and

  $z_2=2\sqrt[3]{2}\cdot e^{i\cdot \left(\frac{14\pi }{9}\right)}$ .

Explanation of Solution

Given information:

The complex number, $-8+8\sqrt{3}i$

To express $-8+8\sqrt{3}i$ in exponential form $z=r{e}^{\theta }$ , find $r$ and the argument $\theta$ .

Comparing $z=-8+8\sqrt{3}i$ with $z=x+iy$ ,

  $\Rightarrow x=-8$ and $y=8\sqrt{3}$

Substitute the values of $x=-8$ and $y=8\sqrt{3}$ in the formula $r=\sqrt{x^2+y^2}$

  $r=\sqrt{{\left(-8\right)}^2+{\left(8\sqrt{3}\right)}^2}=\sqrt{64+192}=\sqrt{256}=16$

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

  $\Rightarrow \sin \theta =\frac{8\sqrt{3}}{16}=\frac{\sqrt{3}}{2}$ and $\cos \theta =\frac{-8}{16}=-\frac{1}{2}$

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

Therefore, the exponential form of a complex number $z=i$ is $z=16\cdot e^{i\cdot \frac{2\pi }{3}}$ .

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 cube roots of $z=16\cdot e^{i\cdot \frac{2\pi }{3}}$ by using the above formula,

The cube roots of $z=16\cdot e^{i\cdot \frac{2\pi }{3}}$ are as below,

  $z_k=2\sqrt[3]{2}\left[e^{i\cdot \left(\frac{2\pi }{9}+\frac{6\pi k}{9}\right)}\right]$ where $k=0,1,2$

For $k=0$ ,

  $z_0=2\sqrt[3]{2}\left[e^{i\cdot \left(\frac{2\pi }{9}+\frac{6\pi \left(0\right)}{9}\right)}\right]=2\sqrt[3]{2}\cdot e^{i\cdot \left(\frac{2\pi }{9}\right)}$

For $k=1$ ,

  $z_1=2\sqrt[3]{2}\left[e^{i\cdot \left(\frac{2\pi }{9}+\frac{6\pi \left(1\right)}{9}\right)}\right]=2\sqrt[3]{2}\cdot e^{i\cdot \left(\frac{8\pi }{9}\right)}$

For $k=2$ ,

  $z_2=2\sqrt[3]{2}\left[e^{i\cdot \left(\frac{2\pi }{9}+\frac{6\pi \left(2\right)}{9}\right)}\right]=2\sqrt[3]{2}\cdot e^{i\cdot \left(\frac{14\pi }{9}\right)}$

Therefore, the complex cube roots of $-8+8\sqrt{3}i$ are $z_0=2\sqrt[3]{2}\cdot e^{i\cdot \left(\frac{2\pi }{9}\right)}$ , $z_1=2\sqrt[3]{2}\cdot e^{i\cdot \left(\frac{8\pi }{9}\right)}$ , and

  $z_2=2\sqrt[3]{2}\cdot e^{i\cdot \left(\frac{14\pi }{9}\right)}$ .

The graph showing the complex cube roots of $-8+8\sqrt{3}i$ is as below.