Proof $\frac{|z_1|}{|z_2|}=\left| \frac{z_1}{z_2}\right|$

Question

Prove: $\frac{|z_1|}{|z_2|}=\left| \frac{z_1}{z_2}\right|$

$$\frac{|z_1|}{|z_2|}=\frac{|a+bi|}{|c+di|}=\frac{\sqrt{a^2+b^2}}{\sqrt{c^2+d^2}}=\sqrt{\frac{{a^2+b^2}}{{c^2+d^2}}}=\left|\frac{z_1}{z_2}\right| $$

I am not sure about the last part $$\sqrt{\frac{{a^2+b^2}}{{c^2+d^2}}}=\left|\frac{z_1}{z_2}\right| $$

Answer 1

To finish off, $$ \frac{z_1}{z_2} = \frac{a+ib}{c+id} = \frac{(a+ib)(c-id)}{c^2+d^2} = \frac{(ac+bd)+i(bc-ad)}{c^2+d^2} $$ so \begin{align} \left|\frac{z_1}{z_2}\right| &= \frac{\sqrt{(ac+bd)^2+(bc-ad)^2}}{c^2+d^2} \\ &= \frac{\sqrt{a^2c^2 + a^2d^2 + b^2c^2 + b^2d^2}}{c^2+d^2} = \frac{\sqrt{(a^2+b^2)(c^2+d^2)}}{c^2+d^2} = \frac{\sqrt{a^2+b^2}}{\sqrt{c^2+d^2}}. \end{align}

(There are quicker ways to do it if you already know that $|zw| = |z|\,|w|$.)

Answer 2

No, the last part is missing: $|\frac{z_1}{z_2} | = \frac{|z_1 (c-di)|}{c^2+d^2} = \frac{|ac+bd+i(bc-ad)|}{|z_2|^2} = ...$

Answer 3

Write $z_1=a+di$ and $z_2=c+di$ then \begin{align*} \frac{z_1}{z_2}&=\frac{a+bi}{c+di}\\ &\frac{a+bi}{c+di}\frac{c-di}{c-di}\\ &\frac{ac+bd+i(bc-ad)}{c^2+d^2}\\ \end{align*} Then $$ |\frac{z_1}{z_2}|^2=\frac{(ac+bd)^2+(bc-ad)^2}{(c^2+d^2)^2}=\frac{a^2+b^2}{c^2+d^2} $$