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.
Best Answer
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