$$ \mid{\frac{e^{-jA}}{B+jC}} \mid =\frac{1}{\sqrt{(B^2+C^2)}} $$
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?
Best Answer
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...