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?

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 $\arg\left\{\dfrac{a+\mathrm{i} \ b}{c+\mathrm{i} \ d}\right\}=\arg\left\{\frac{\rho_1}{\rho_2}e^{\mathrm{i}(\varphi_1-\varphi_2)}\right\}=\varphi_1-\varphi_2$.
