Chapter 8, Problem 4T

(a)

To determine

To sketch: The point $z=1+\sqrt{3}i$ in complex plane.

Explanation of Solution

The complex number is $z=1+\sqrt{3}i$.

The real part of the number is 1 and the imaginary part is $\sqrt{3}$.

The graph of the complex number is,

Figure (1)

The above graph shows the point $z=1+\sqrt{3}i$ in complex plane.

(b)

To determine

The polar form of the given complex number.

Answer to Problem 4T

The polar form of the complex number $z=1+\sqrt{3}i$ is $z=2\left(\cos \frac{\pi }{3}+i\sin \frac{\pi }{3}\right)$.

Explanation of Solution

Given:

The complex number is $z=1+\sqrt{3}i$.

Calculation:

The modulus r of the complex number $z=a+ib$ is,

$r=\sqrt{a^2+b^2}$

Substitute 1 for a and $\sqrt{3}$ for b in the above expression.

$\begin{array}{c}r=\sqrt{{\left(1\right)}^2+{\left(\sqrt{3}\right)}^2}\\ =\sqrt{1+3}\\ =2\end{array}$

The modulus of the complex number is 2.

The argument $\theta$ of the complex number $z=a+ib$ is,

$\begin{array}{c}\tan \theta =\frac{b}{a}\\ \theta ={\tan }^{-1}\left(\frac{b}{a}\right)\end{array}$

Substitute 1 for a and $\sqrt{3}$ for b in the above expression.

$\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{\sqrt{3}}{1}\right)\\ =\frac{\pi }{3}\end{array}$

The argument $\theta$ of the complex number $z=1+\sqrt{3}i$ is $\frac{\pi }{3}$.

The given complex number is $z=1+\sqrt{3}i$.

In general the polar form of the complex number $z=a+ib$ is,

$z=r\left(\cos \theta +i\sin \theta \right)$ (1)

The value of r is 2 and the value of $\theta$ is $\frac{\pi }{3}$.

Substitute 2 for r and $\frac{\pi }{3}$ for $\theta$ in equation (1) to get the polar form.

$z=2\left(\cos \frac{\pi }{3}+i\sin \frac{\pi }{3}\right)$

Thus, the polar form of the complex number $z=1+\sqrt{3}i$ is $z=2\left(\cos \frac{\pi }{3}+i\sin \frac{\pi }{3}\right)$.

(c)

To determine

The value of $z^9={\left(1+\sqrt{3}i\right)}^9$.

Answer to Problem 4T

The value of ${\left(1+\sqrt{3}i\right)}^9$ is $-512$.

Explanation of Solution

Given:

The below number is a complex number with power 9.

${\left(1+\sqrt{3}i\right)}^9$

Formula used:

De Moivre’s Theorem.

$z^n=r^n\left(\cos n\theta +i\sin n\theta \right)$ (2)

Where,

The number $n$ is an integer.

Calculation:

From part (b) the polar form of the complex number $z=1+\sqrt{3}i$ is,

$z=2\left(\cos \frac{\pi }{3}+i\sin \frac{\pi }{3}\right)$

Substitute 9 for $n$, $2$ for $r$ and $\frac{\pi }{3}$ for $\theta$ in the equation (2) to find the value of ${\left(1+\sqrt{3}i\right)}^9$

$\begin{array}{c}{z}^9=2^9\left(\cos \left(9\times \frac{\pi }{3}\right)+i\sin \left(9\times \frac{\pi }{3}\right)\right)\\ =512\left(\cos 3\pi +i\sin 3\pi \right)\\ =512\left(-1+0\right)\left(\because \cos 3\pi =-1,\sin 3\pi =0\right)\\ =-512\end{array}$

Thus, the value of ${\left(1+\sqrt{3}i\right)}^9$ is $-512$.