# Prove $|z_1/z_2| = |z_1|/|z_2|$ without using the polar form

Prove $|z_1/z_2| = |z_1|/|z_2|$ without using the polar form.

Let $z_1 = a_1+b_1i$, $z_2 = a_2+b_2i$. Then $|z_1| = \sqrt{a_1^2+b_1^2}$ and $|z_2| = \sqrt{a_2^2+b_2^2}$. Hence $$RHS = \frac{|z_1|}{|z_2|} = \frac{\sqrt{a_1^2+b_1^2}}{\sqrt{a_2^2+b_2^2}}$$ $$LHS = \left|\frac{z_1}{z_2}\right| = \left|\frac{(a_1+b_1i)(a_2-b_2i)}{a_2^2-b_2^2}\right| = \left|\frac{a_1a_2+b_1b_2}{a_2^2-b_2^2} + i\frac{a_2b_1 - a_1b_2}{a_2^2-b_2^2}\right| \\= \frac{\sqrt{a_1^2a_2^2+b_1^2b_2^2+a_2^2b_1^2-a_1^2b_2^2}}{a_2^2-b_2^2}.$$ I'm stuck. Help please.

No polar because this is presented in the context of absolute value and before polar form.

• Hint : use polar form. – C. Dubussy Jul 12 '16 at 22:14
• @C.Dubussy This part is discuss before polar form – user198044 Jul 12 '16 at 22:18

## 4 Answers

You had some wrong signs

\begin{align*}|\frac{z_1}{z_2}| & = |\frac{(a_1+b_1i)(a_2-b_2i)}{a_2^2\color{red}{+}b_2^2}|\\ & = |\frac{a_1a_2+b_1b_2}{a_2^2+b_2^2} + i\frac{a_2b_1 - a_1b_2}{a_2^2+b_2^2}|\\ & = \frac{\sqrt{a_1^2a_2^2+b_1^2b_2^2+a_2^2b_1^2\color{red}{+}a_1^2b_2^2}}{a_2^2+b_2^2} \\ & =\frac{\sqrt{(a_1^2+b_1^2)(a_2^2+b_2^2)}}{a_2^2+b_2^2} \\ & =\frac{\sqrt{a_1^2+b_1^2}}{\sqrt{a_2^2+b_2^2}} \end{align*} As claimed

$|\frac{1}{z}|=\frac{1}{|z|}$

in fact : $z=a+ib$ so $|z|^2=a^2+b^2$ and $\frac{1}{z}=\frac{1}{a+ib}=\frac{a-ib}{(a-ib)(a+ib)}=\frac{a-ib}{a^2+b^2}=\frac{1}{a^2+b^2}(a-ib)$ then : $$|\frac{1}{z}|^2= (\frac{1}{a^2+b^2})^2(a^2+b^2)=\frac{1}{a^2+b^2}=\frac{1}{|z|^2}$$

|zw|=|z||w|

$z=a+ib$ and $w=c+id$ so $|z|^2=a^2+b^2$ and $|w|^2=c^2+d^2$ $zw=(a+ib)(c+id)=ac-bd+i(ac+bd)$ then : $$|zw|^2= (ac-bd)^2+(bc+ad)^2= (ac)^2+(db)^2+(bc)^2+(ad)^2=a^2(c^2+d^2)+b^2(c^2+d^2)=(a^2+b^2)(c^2+d^2)=|z|^2|w|^2$$

Concerning the left hand side, you may rather write \begin{align} \left|\frac{z_1}{z_2}\right|&=\left|\frac{(a_1+b_1i)(a_2-b_2i)}{a_2^2+b_2^2}\right| \\\\&=\frac{\sqrt{(a_1a_2+b_1b_2)^2+(a_1b_2-a_2b_1)^2}}{a_2^2+b_2^2} \\\\&=\frac{\sqrt{a^2_1a^2_2+a_1^2b^2_2+a^2_2b^2_1+b^2_1b^2_2}}{a_2^2+b_2^2} \\\\&=\frac{\sqrt{(a^2_1+b^2_1)(a^2_2+b^2_2)}}{a_2^2+b_2^2} \\\\&=\frac{\sqrt{a_1^2+b_1^2}}{\sqrt{a_2^2+b_2^2}}. \end{align}

Check your signs on your last line...

$\dfrac {|z_1 \bar z_2|}{|z_2\bar z_2|} = \dfrac {\sqrt{a_1^2a_2^2 + a_1^2b_2^2+a_2^2b_1^2 + b_1^2b_2^2}}{a_2^2+b_2^2}\\ \dfrac {\sqrt{a_1^2(a_2^2 +b_2^2) + b_1^2(a_2^2 + b_2^2)}}{a_2^2+b_2^2}\\ \dfrac {\sqrt{(a_1^2 + b_1^2)(a_2^2 +b_2^2)}}{a_2^2+b_2^2}\\ \dfrac {|z_1||z_2|}{|z_2|^2}\\ \dfrac {|z_1|}{|z_2|}$