Chapter 6.6, Problem 30E

Solution

a)

To find: The product $z_1\cdot z_2$ and quotient $z_1/z_2$ using the polar form for $z_1$ and $z_2$ .

The product is $-3.20-6.46i$ and quotient is $1.80+0.12i$ .

Given information:

It is given that $z_1=2-3i$ and $z_2=1-\sqrt{3}i$ .

Formula used:

The product of two complex numbers is given by,

  $z_1\cdot z_2=r_1{r}_2\left[\cos \left({\theta }_1+{\theta }_2\right)+i\sin \left({\theta }_1+{\theta }_2\right)\right]$

The quotient of two complex numbers is given by,

  $\frac{z_1}{z_2}=\frac{r_1}{r_2}\left[\cos \left({\theta }_1-{\theta }_2\right)+i\sin \left({\theta }_1-{\theta }_2\right)\right]$

Calculation:

Consider the first complex number.

  $z_1=2-3i$

From the given complex number,

  $\begin{array}{c}a=2\\ b=-3\end{array}$

Substitute $2$ for $a$ and $-3$ for $b$ in equation $r=\sqrt{a^2+b^2}$ to find modulus $r$ .

  $\begin{array}{c}r=\sqrt{{\left(2\right)}^2+{\left(-3\right)}^2}\\ =\sqrt{4+9}\\ =\sqrt{13}\end{array}$

Substitute $2$ for $a$ and $-3$ for $b$ in equation $\tan \theta =\frac{b}{a}$ to find argument.

  $\begin{array}{c}\tan \theta =\frac{-3}{2}\\ \theta =2\pi -{\tan }^{-1}\left(\frac{3}{2}\right)\\ =5.30\end{array}$

Substitute $\sqrt{13}$ for $r$ and $5.30$ for $\theta$ in equation $z=r\left(\cos \theta +i\sin \theta \right)$ .

  $z_1=\sqrt{13}\left(\cos 5.30+i\sin 5.30\right)$

Consider the second complex number.

  $z_2=1-\sqrt{3}i$

From the given complex number,

  $\begin{array}{c}a=1\\ b=-\sqrt{3}\end{array}$

Substitute $1$ for $a$ and $-\sqrt{3}$ for $b$ in equation $r=\sqrt{a^2+b^2}$ to find modulus $r$ .

  $\begin{array}{c}r=\sqrt{{\left(1\right)}^2+{\left(-\sqrt{3}\right)}^2}\\ =\sqrt{1+3}\\ =\sqrt{4}\\ =2\end{array}$

Substitute $1$ for $a$ and $-\sqrt{3}$ for $b$ in equation $\tan \theta =\frac{b}{a}$ to find argument.

  $\begin{array}{c}\tan \theta =\frac{-\sqrt{3}}{1}\\ \theta =2\pi -{\tan }^{-1}\left(\sqrt{3}\right)\\ =\frac{5\pi }{3}\end{array}$

Substitute $2$ for $r$ and $\frac{5\pi }{3}$ for $\theta$ in equation $z=r\left(\cos \theta +i\sin \theta \right)$ .

  $z_2=2\left(\cos \frac{5\pi }{3}+i\sin \frac{5\pi }{3}\right)$

Find the product of two complex numbers.

  $\begin{array}{c}{z}_1\cdot z_2=\left(\sqrt{13}\right)\left(2\right)\left(\cos \left(5.30+\frac{5\pi }{3}\right)+i\sin \left(5.30+\frac{5\pi }{3}\right)\right)\\ =2\sqrt{13}\left(\cos 10.536+i\sin 10.536\right)\\ =-3.20-6.46i\end{array}$

Find the quotient of two complex numbers.

  $\begin{array}{c}\frac{z_1}{z_2}=\frac{\sqrt{13}}{2}\left(\cos \left(5.30-\frac{5\pi }{3}\right)+i\sin \left(5.30-\frac{5\pi }{3}\right)\right)\\ =\frac{\sqrt{13}}{2}\left(\cos 0.064+i\sin 0.064\right)\\ =1.80+0.12i\end{array}$

Therefore, the product is $-3.20-6.46i$ and quotient is $1.80+0.12i$ .

b)

To find: The product $z_1\cdot z_2$ and quotient $z_1/z_2$ using the standardform for $z_1$ and $z_2$ .

The product is $-3.20-6.46i$ and quotient is $1.80+0.12i$ .

Given Information:

It is given that $z_1=2-3i$ and $z_2=1-\sqrt{3}i$ .

Calculation:

Find the product of two complex numbers.

  $\begin{array}{c}{z}_1\cdot z_2=\left(2-3i\right)\left(1-\sqrt{3}i\right)\\ =\left(2\right)\left(1\right)+\left(2\right)\left(-\sqrt{3}i\right)+\left(-3i\right)\left(1\right)+\left(-3i\right)\left(-\sqrt{3}i\right)\\ =2-2\sqrt{3}i-3i+3\sqrt{3}{i}^2\\ =2-2\sqrt{3}i-3i+3\sqrt{3}\left(-1\right)\end{array}$

Simplify further.

  $\begin{array}{c}{z}_1\cdot z_2=\left(2-3\sqrt{3}\right)-\left(3+2\sqrt{3}\right)i\\ =-3.20-6.46i\end{array}$

Find the quotient of two complex numbers.

  $\begin{array}{c}\frac{z_1}{z_2}=\frac{1-i}{1-\sqrt{3}i}\cdot \frac{1+\sqrt{3}i}{1+\sqrt{3}i}\\ =\frac{\left(2\right)\left(1\right)+\left(2\right)\left(\sqrt{3}i\right)+\left(-3i\right)\left(1\right)+\left(-3i\right)\left(\sqrt{3}i\right)}{1^2-{\left(\sqrt{3}i\right)}^2}\\ =\frac{2+2\sqrt{3}i-3i-3\sqrt{3}\left(-1\right)}{1-3\left(-1\right)}\\ =\frac{\left(2+3\sqrt{3}\right)-\left(3-2\sqrt{3}\right)i}{4}\end{array}$

Simplify further.

  $\begin{array}{c}\frac{z_1}{z_2}=\frac{2+3\sqrt{3}}{4}-\frac{3-2\sqrt{3}}{4}i\\ =1.80+0.12i\end{array}$

Therefore, the product is $-3.20-6.46i$ and quotient is $1.80+0.12i$ .