Chapter 6.6, Problem 72E

Solution

a)

To prove: That $\left|z\right|=\left|r\right|$ .

Formula used:

By distributive property,

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

The modulus of $a+ib$ is $\sqrt{a^2+b^2}$ . Here, $a=r\cos \theta$ and $b=r\sin \theta$ .

Calculation:

Find the $\left|z\right|$ as follows.

  $\begin{array}{c}\left|z\right|=\sqrt{{\left(r\cos \theta \right)}^2+{\left(r\sin \theta \right)}^2}\\ =\sqrt{r^2{\cos }^2\theta +r^2{\sin }^2\theta }\\ =\sqrt{r^2\left({\cos }^2\theta +{\sin }^2\theta \right)}\\ =\sqrt{r^2\left(1\right)}\end{array}$

Simplify further.

  $\begin{array}{c}\left|z\right|=\sqrt{r^2}\\ =\left|r\right|\end{array}$

Therefore, it is proved that $\left|z\right|=\left|r\right|$ .

b)

To prove: That $\left|z_1\cdot z_2\right|=\left|z_1\right|\cdot \left|z_2\right|$ .

Formula used:

The product of two complex numbers is given by,

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

Calculation:

Let,

  $\begin{array}{c}{z}_1=r_1\left(\cos {\theta }_1+i\sin {\theta }_2\right)\\ z_2=r_2\left(\cos {\theta }_2+i\sin {\theta }_2\right)\end{array}$

Find the product of two complex numbers.

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

It is known that $\left|z_1\cdot z_2\right|=\left|r_1\cdot r_2\right|$

Use the property $\left|ab\right|=\left|a\right|\left|b\right|$ .

  $\left|z_1\cdot z_2\right|=\left|r_1\right|\cdot \left|r_2\right|$

The right side is the product of the moduli of $z_1$ and $z_2$ so by substitution,

  $\left|z_1\cdot z_2\right|=\left|z_1\right|\cdot \left|z_2\right|$

Therefore, it is proved that $\left|z_1\cdot z_2\right|=\left|z_1\right|\cdot \left|z_2\right|$ .