Chapter 6.6, Problem 27E

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 $5+i$ and quotient is $\frac{1}{2}-\frac{5}{2}i$ .

Given information:

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

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.

  $z_1=3-2i$

From the given complex number,

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

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

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

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

  $\begin{array}{c}\tan \theta =\frac{-2}{3}\\ \theta ={\tan }^{-1}\left(\frac{-2}{3}\right)\\ =2\pi -{\tan }^{-1}\left(\frac{2}{3}\right)\\ =5.70\end{array}$

Substitute $\sqrt{13}$ for $r$ and $5.70$ for $\theta$ in equation $z=r\left(\cos \theta +i\sin \theta \right)$ .

  $z_1=\sqrt{13}\left(\cos 5.70+i\sin 5.70\right)$

Consider the second complex number.

  $z_2=1+i$

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 $r$ .

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

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

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

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

  $z_2=\sqrt{2}\left(\cos \frac{\pi }{4}+i\sin \frac{\pi }{4}\right)$

Find the product of two complex numbers.

  $\begin{array}{c}{z}_1\cdot z_2=\left(\sqrt{13}\right)\left(\sqrt{2}\right)\left(\cos \left(5.70+\frac{\pi }{4}\right)+i\sin \left(5.70+\frac{\pi }{4}\right)\right)\\ =\sqrt{26}\left(\cos 6.48+i\sin 6.48\right)\\ =5+i\end{array}$

Find the quotient of two complex numbers.

  $\begin{array}{c}\frac{z_1}{z_2}=\frac{\left(\sqrt{13}\right)}{\left(\sqrt{2}\right)}\left(\cos \left(5.70-\frac{\pi }{4}\right)+i\sin \left(5.70-\frac{\pi }{4}\right)\right)\\ =\sqrt{6.5}\left(\cos 4.91+i\sin 4.91\right)\\ =0.5-2.5i\\ =\frac{1}{2}-\frac{5}{2}i\end{array}$

Therefore, the product is $5+i$ and quotient is $\frac{1}{2}-\frac{5}{2}i$ .

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 $5+i$ and quotient is $\frac{1}{2}-\frac{5}{2}i$ .

Given information:

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

Calculation:

Find the product of two complex numbers.

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

Simplify further.

  $\begin{array}{c}{z}_1\cdot z_2=3+3i-2i+2\\ =5+i\end{array}$

Find the quotient of two complex numbers.

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

Simplify further.

  $\frac{z_1}{z_2}=\frac{1}{2}-\frac{5}{2}i$

Therefore, the product is $5+i$ and quotient is $\frac{1}{2}-\frac{5}{2}i$ .