Is this a valid method of finding magnitude of complex fraction If I have a complex fraction $\dfrac{a+bi}{c+di}$ and I want the magnitude, then will it be $\left|\dfrac{a+bi}{c+di}\right|=\dfrac{|a+bi|}{|c+di|}$?
Scratch that ... I just found the answer on another page; however, I'm still unclear why it's true?
 A: $\frac{a+bi}{c+di} = \frac{a+bi}{c+di} * \frac{c-di}{c-di} = i (\frac{b c}{c^2+d^2}-\frac{a d}{c^2+d^2})+\frac{a c}{c^2+d^2}+\frac{b d}{c^2+d^2}$. At this point, you should be able to get the magnitude easily. Yes, it'll be cumbersome computation wise, but that should be it. 
Suppose $e = \frac{b c}{c^2+d^2}-\frac{a d}{c^2+d^2}$ and $f = \frac{a c}{c^2+d^2}+\frac{b d}{c^2+d^2}$
Then, $\|f + ei\| = \sqrt{f^2+e^2} = \sqrt{\frac{(bc-ad)^2}{(c^2+d^2)^2} + \frac{(ac+bd)^2}{(c^2+d^2)^2}} = \sqrt{\frac{2(a^2d^2+b^2c^2)}{(c^2+d^2)^2}}$ and you could take it from there.
A: You can make use of complex exponents.
$$\dfrac{a+\mathrm{i} \ b}{c+\mathrm{i} \ d}=\frac{\rho_1e^{\mathrm{i} \varphi_1}}{\rho_2e^{\mathrm{i} \varphi_2}}=\frac{\rho_1}{\rho_2}e^{\mathrm{i}(\varphi_1-\varphi_2)}$$
where $\rho_1=\sqrt{a^2+b^2}, \rho_2=\sqrt{c^2+d^2}$ are the magnitudes and $\varphi_1=\arg\{a+\mathrm{i} \ b\},\varphi_2=\arg\{c+\mathrm{i} \ d\}$ are phases of $a+\mathrm{i} \ b$ and $c+\mathrm{i} \ d$ respectively. 
Then since $\rho_1, \rho_2$ are real (and positive) and the absolute value of complex exponent is $1$:
 $$\left| \dfrac{a+\mathrm{i} \ b}{c+\mathrm{i} \ d}\right|=\left|\frac{\rho_1}{\rho_2}e^{\mathrm{i}(\varphi_1-\varphi_2)} \right|=\left|\frac{\rho_1}{\rho_2}\right|\left|e^{\mathrm{i}(\varphi_1-\varphi_2)} \right|=\left|\frac{\rho_1}{\rho_2}\right|=\frac{\left|\rho_1\right|}{\left|\rho_2\right|}=\frac{\left|a+\mathrm{i} \ b\right|}{\left|c+\mathrm{i} \ d\right|}.$$
Moreover, using complex exponents it is easy to show that $$\arg\left\{\dfrac{a+\mathrm{i} \ b}{c+\mathrm{i} \ d}\right\}=\arg\left\{a+\mathrm{i} \ b\right\}-\arg\left\{c+\mathrm{i} \ d\right\}.$$
That is true, since $\arg\left\{\dfrac{a+\mathrm{i} \ b}{c+\mathrm{i} \ d}\right\}=\arg\left\{\frac{\rho_1}{\rho_2}e^{\mathrm{i}(\varphi_1-\varphi_2)}\right\}=\varphi_1-\varphi_2$.
A: A simpler approach:
Let $z_1=a+bi$ and $z_2=c+di$.  Since by properties of absolute value we have $|z_1z_2|=|z_1||z_2|,$ and the fact that $z_2(\frac{z_1}{z_2})=z_1$ then we have that $$\left|z_2\frac{z_1}{z_2}\right|=|z_1|\implies|z_2|\bigg|\frac{z_1}{z_2}\bigg|=|z_1|\implies \bigg|\frac{z_1}{z_2}\bigg|=\frac{|z_1|}{|z_2|}$$
