Q:

Absolute value of complex exponential divided by complex number

$$ \mid{\frac{e^{-jA}}{B+jC}} \mid =\frac{1}{\sqrt{(B^2+C^2)}} $$
It is clear for me why $ \mid e^{-jA} \mid =1$. I just need $e^{-jA}$ to represent like $\cos A+j\sin B$ and take absolute value.
I cant understand the final solution.  Where is $B-jC$?
My solution
If I take absolute value from $ \mid{\frac{e^{-jA}}{B+jC}}\mid$ then i should :
$ {\frac{e^{-jA}}{B+jC}}\frac{B-jC}{B-jC} = {\frac{e^{-jA}}{B^2+C^2}}(B-jC)$
Can someone explain to me why my solution is wrong?

A:

There are two complex operations one is the modulus and one is conjugate. The modulus operator has the following property:
$$\mid \frac{z_1}{z_2}\mid=\frac{\mid z_1\mid}{\mid z_2\mid}$$For your problem:
$$\mid \frac{e^{-jA}}{B-jC} \mid=\frac{\mid e^{-jA} \mid}{\mid B-jC \mid}=\frac{1}{\mid B-jC \mid}=\frac{1}{\sqrt{B^2+C^2}}$$
The problem with your solution was that :
$$\mid\frac{e^{-jA}(B-jC)}{B^2+C^2}\mid=\frac{\mid e^{-jA}(B-jC) \mid}{\mid B^2+C^2 \mid}=\frac{\mid e^{-jA}\mid\mid(B-jC) \mid}{\mid(B^2+C^2) \mid}$$
$$\frac{1(\sqrt{B^2+C^2})}{B^2+C^2}=\frac{1}{\sqrt{B^2+C^2}}$$
Hope this helps...

