Chapter 6.6, Problem 29E

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 $18-4i$ and quotient is $0.35-41i$ .

Given information:

It is given that $z_1=3+i$ and $z_2=5-3i$ .

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.

  $z_1=3+i$

From the given complex number,

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

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

  $\begin{array}{c}r=\sqrt{{\left(3\right)}^2+{\left(1\right)}^2}\\ =\sqrt{10}\end{array}$

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

  $\begin{array}{c}\tan \theta =\frac{1}{3}\\ \theta ={\tan }^{-1}\left(\frac{1}{3}\right)\\ =0.32\end{array}$

Substitute $2$ for $r$ and $0.32$ for $\theta$ in equation $z=r\left(\cos \theta +i\sin \theta \right)$ .

  $z_1=2\left(\cos 0.32+i\sin 0.32\right)$

Consider the second complex number.

  $z_2=5-3i$

From the given complex number,

  $\begin{array}{c}a=5\\ b=-3\end{array}$

Substitute $5$ for $a$ and $-3$ for $b$ in equation $r=\sqrt{a^2+b^2}$ to find modulus $r$ .

  $\begin{array}{c}r=\sqrt{{\left(5\right)}^2+{\left(-3\right)}^2}\\ =\sqrt{25+9}\\ =\sqrt{34}\end{array}$

Substitute $5$ for $a$ and $-3$ for $b$ in equation $\tan \theta =\frac{b}{a}$ to find argument.

  $\begin{array}{c}\tan \theta =\frac{-3}{5}\\ \theta =2\pi -{\tan }^{-1}\left(\frac{3}{5}\right)\\ =5.74\end{array}$

Substitute $\sqrt{34}$ for $r$ and $5.74$ for $\theta$ in equation $z=r\left(\cos \theta +i\sin \theta \right)$ .

  $z_2=\sqrt{34}\left(\cos 5.74+i\sin 5.74\right)$

Find the product of two complex numbers.

  $\begin{array}{c}{z}_1\cdot z_2=\left(\sqrt{10}\right)\left(\sqrt{34}\right)\left(\cos \left(0.32+5.74\right)+i\sin \left(0.32+5.74\right)\right)\\ =\sqrt{340}\left(\cos 6.06+i\sin 6.06\right)\\ =18-4i\end{array}$

Find the quotient of two complex numbers.

  $\begin{array}{c}\frac{z_1}{z_2}=\frac{\sqrt{10}}{\sqrt{34}}\left(\cos \left(0.32-5.74\right)+i\sin \left(0.32-5.74\right)\right)\\ =\sqrt{\frac{5}{17}}\left(\cos \left(-5.42\right)+i\sin \left(-5.42\right)\right)\\ =0.35+41i\end{array}$

Therefore, the product is $18-4i$ and quotient is $0.35-41i$ .

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 $18-4i$ and quotient is $0.35-41i$ .

Given Information:

It is given that $z_1=3+i$ and $z_2=5-3i$ .

Calculation:

Find the product of two complex numbers.

  $\begin{array}{c}{z}_1\cdot z_2=\left(3+i\right)\left(5-3i\right)\\ =\left(3\right)\left(5\right)+\left(3\right)\left(-3i\right)+\left(i\right)\left(5\right)+\left(i\right)\left(-3i\right)\\ =15-9i+5i-3i^2\\ =15-4i-3\left(-1\right)\end{array}$

Simplify further.

  $\begin{array}{c}{z}_1\cdot z_2=15-4i+3\\ =18-4i\end{array}$

Find the quotient of two complex numbers.

  $\begin{array}{c}\frac{z_1}{z_2}=\frac{3+i}{5-3i}\cdot \frac{5+3i}{5+3i}\\ =\frac{\left(3\right)\left(5\right)+\left(3\right)\left(3i\right)+\left(i\right)\left(5\right)+\left(i\right)\left(3i\right)}{5^2-{\left(3i\right)}^2}\\ =\frac{15+9i+5i+3i^2}{25-9i^2}\\ =\frac{15+14i+3\left(-1\right)}{25-9\left(-1\right)}\end{array}$

Simplify further.

  $\begin{array}{c}\frac{z_1}{z_2}=\frac{12+14i}{34}\\ =\frac{6}{17}+\frac{7}{17}i\\ =0.35+0.41i\end{array}$

Therefore, the product is $18-4i$ and quotient is $0.35-41i$ .