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 $$ |\frac{z_1}{z_2}|^2=\frac{(ac+bd)^2+(bc-ad)^2}{(c^2+d^2)^2}=\frac{a^2+b^2}{c^2+d^2} $$

Answer 4

Suppose $z_1=x_1+iy_1$ and $z_2=x_2+iy_2$ $$ \begin{align} \text{LHS}\Rightarrow\frac{\left| z_1 \right| }{\left| z_2 \right|} &=\frac{\left| x_1+iy_1 \right| }{\left| x_2+iy_2 \right|}\\ &=\frac{\sqrt{x_1^2+y_1^2} }{\sqrt{x_2^2+y_2^2}} \\ &=\frac{\sqrt{(x_1^2+y_1^2)(x_2^2+y_2^2)} }{x_2^2+y_2^2} \ (1) \end{align} $$

$$ \begin{align} \text{RHS}\Rightarrow \left| \frac{z_1}{z_2} \right| &=\left| \frac{x_1+iy_1}{x_2+iy_2} \right|\\ \\ &=\left| \frac{x_1+iy_1}{x_2+iy_2} \cdot \frac{x_2-iy_2}{x_2-iy_2} \right|\\ \\ &=\left| \frac{(x_1+iy_1)(x_2-iy_2)}{(x_2+iy_2)(x_2-iy_2)}\right|\\ \\ &=\left| \frac{(x_1x_2+y_1y_2)+i(-x_1y_2+x_2y_1)}{x_2^2+y_2^2}\right|\\ \\ &= \frac{\sqrt{(x_1x_2+y_1y_2)^2+(-x_1y_2+x_2y_1)^2}}{x_2^2+y_2^2}\\ &= \frac{\sqrt{(x_1^2x_2^2+y_1^2y_2^2+2 x_1x_2y_1y_2)+(x_1^2x_2^2+y_1^2y_2^2-2 x_1x_2y_1y_2)}}{x_2^2+y_2^2}\\ \\ &= \frac{\sqrt{x_1^2x_2^2+y_1^2y_2^2+x_1^2x_2^2+y_1^2y_2^2}}{x_2^2+y_2^2}\\ \\ &= \frac{\sqrt{\\ \\ x_1^2\Big(x_2^2+y_2^2\Big)+y_1^2\Big(x_2^2+y_2^2\Big)}}{x_2^2+y_2^2}\\ \\ &=\frac{\sqrt{(x_1^2+y_1^2)(x_2^2+y_2^2)} }{x_2^2+y_2^2} \ (2) \end{align} $$

See, (1) and (2) are equivalent.

Note: Fraction $\frac{\sqrt{(x_1^2+y_1^2)(x_2^2+y_2^2)} }{x_2^2+y_2^2}$ for Non-Zero values of $x_{1,2}$ and $y_{1,2}$ has answer.

Answer 5

We can simplify things somewhat by breaking the problem into two parts $|z_1 z_2|=|z_1||z_2|$, and $|1/z|=1/|z|$.

For the first part, $|z_1 z_2|=\sqrt{z_1 z_2 \overline{z_1 z_2}}=\sqrt{z_1\overline{z_1}z_2\overline{z_2}}=\sqrt{z_1\overline{z_1}}\sqrt{z_2\overline{z_2}}=|z_1||z_2|$.

For the second part, $|1/z|=|\overline{z}/(z\overline z)|=|\overline{z}/|z|^2|=|z|/|z|^2$, where we have pulled out the real number $1/|z|^2$ from the expression and used the fact that $|z|=|\overline{z}|$.

Answer 6

Just multiply the denominator with conjugate then simplify.. You will get the answer