Chapter 6.6, Problem 99E

To determine

To prove:The result on quotient of two complex numbers. That is

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

Explanation of Solution

Giveninformation:

The two complex numbers are $z_1=r_1(\cos {\theta }_1+i\sin {\theta }_1)$ and $z_2=r_2(\cos {\theta }_2+i\sin {\theta }_2)$ , where $z_2\ne 0$ .

Proof:

Let $z_1=r_1(\cos {\theta }_1+i\sin {\theta }_1)$ and $z_2=r_2(\cos {\theta }_2+i\sin {\theta }_2)$ are two complex numbers, then their quotient is

  $\frac{z_1}{z_2}=\frac{r_1(\cos {\theta }_1+i\sin {\theta }_1)}{r_2(\cos {\theta }_2+i\sin {\theta }_2)}$ .

Rationalizing the denominator, we have

  $\begin{array}{l}\frac{z_1}{z_2}=\frac{r_1(\cos {\theta }_1+i\sin {\theta }_1)}{r_2(\cos {\theta }_2+i\sin {\theta }_2)}\times \frac{(\cos {\theta }_2-i\sin {\theta }_2)}{(\cos {\theta }_2-i\sin {\theta }_2)}\\ =\frac{r_1(\cos {\theta }_1+i\sin {\theta }_1)(\cos {\theta }_2-i\sin {\theta }_2)}{r_2(\cos {\theta }_2+i\sin {\theta }_2)(\cos {\theta }_2-i\sin {\theta }_2)}\\ =\frac{r_1(\cos {\theta }_1\cos {\theta }_2-i\cos {\theta }_1\sin {\theta }_2+i\sin {\theta }_1\cos {\theta }_2-i^2\sin {\theta }_1\sin {\theta }_2)}{r_2({\cos }^2{\theta }_2-i^2{\sin }^2{\theta }_2)}\\ =\frac{r_1(\cos {\theta }_1\cos {\theta }_2+\sin {\theta }_1\sin {\theta }_2+i(\sin {\theta }_1\cos {\theta }_2-\cos {\theta }_1\sin {\theta }_2))}{r_2({\cos }^2{\theta }_2+{\sin }^2{\theta }_2)}\\ =\frac{r_1\cos ({\theta }_1-{\theta }_2)+i\sin ({\theta }_1-{\theta }_2)}{r_2(1)}\\ =\frac{r_1}{r_2}\left[\cos ({\theta }_1-{\theta }_2)+i\sin ({\theta }_1-{\theta }_2)\right]\end{array}$