Chapter 6.6, Problem 28E

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 $2.73-0.73i$ and quotient is $0.18-0.68i$ .

Given information:

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

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.

  $z_1=1-i$

From the given complex number,

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

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

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

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

  $\begin{array}{c}\tan \theta =\frac{-1}{1}\\ \theta ={\tan }^{-1}\left(-1\right)\\ =2\pi -{\tan }^{-1}\left(1\right)\\ =\frac{7\pi }{4}\end{array}$

Substitute $\sqrt{2}$ for $r$ and $\frac{7\pi }{4}$ for $\theta$ in equation $z=r\left(\cos \theta +i\sin \theta \right)$ .

  $z_1=\sqrt{2}\left(\cos \frac{7\pi }{4}+i\sin \frac{7\pi }{4}\right)$

Consider the second complex number.

  $z_2=\sqrt{3}+i$

From the given complex number,

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

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

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

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

  $\begin{array}{c}\tan \theta =\frac{1}{\sqrt{3}}\\ \theta ={\tan }^{-1}\left(\frac{1}{\sqrt{3}}\right)\\ =\frac{\pi }{6}\end{array}$

Substitute $2$ for $r$ and $\frac{\pi }{6}$ for $\theta$ in equation $z=r\left(\cos \theta +i\sin \theta \right)$ .

  $z_2=2\left(\cos \frac{\pi }{6}+i\sin \frac{\pi }{6}\right)$

Find the product of two complex numbers.

  $\begin{array}{c}{z}_1\cdot z_2=\left(\sqrt{2}\right)\left(2\right)\left(\cos \left(\frac{7\pi }{4}+\frac{\pi }{6}\right)+i\sin \left(\frac{7\pi }{4}+\frac{\pi }{6}\right)\right)\\ =2\sqrt{2}\left(\cos \frac{23\pi }{12}+i\sin \frac{23\pi }{12}\right)\\ =2.73-0.73i\end{array}$

Find the quotient of two complex numbers.

  $\begin{array}{c}\frac{z_1}{z_2}=\frac{\sqrt{2}}{2}\left(\cos \left(\frac{7\pi }{4}+\frac{\pi }{6}\right)+i\sin \left(\frac{7\pi }{4}+\frac{\pi }{6}\right)\right)\\ =\frac{\sqrt{2}}{2}\left(\cos \left(\frac{19\pi }{12}\right)+i\sin \left(\frac{19\pi }{12}\right)\right)\\ =0.18-0.68i\end{array}$

Therefore, the product is $2.73-0.73i$ and quotient is $0.18-0.68i$ .

b)

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

The product is $2.73-0.73i$ and quotient is

  $0.18-0.68i$ .

Given information:

It is given that $z_1=3-2i$ and $z_2=1+i$ .

Calculation:

Find the product of two complex numbers.

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

Simplify further.

  $\begin{array}{c}{z}_1\cdot z_2=\left(1+\sqrt{3}\right)+\left(1-\sqrt{3}\right)i\\ =2.73-0.73i\end{array}$

Find the quotient of two complex numbers.

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

Simplify further.

  $\begin{array}{c}\frac{z_1}{z_2}=\frac{\left(\sqrt{3}-1\right)-\left(1+\sqrt{3}\right)i}{4}\\ =\frac{\sqrt{3}-1}{4}-\frac{1+\sqrt{3}}{4}i\\ =0.18-0.68i\end{array}$

Therefore, the product is $2.73-0.73i$ and quotient is $0.18-0.68i$ .