Absolute value of complex exponential divided by complex number

Question

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

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?

Answer 1

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...

Answer 2

I guess $A,B$ and $C$ shall be reals (otherwise the statement is false), hence:

By definition of the absolute value of a complex number it holds $$|B+jC| = \sqrt{B^2 + C^2}$$ and so we get:

$$\left|{\frac{e^{-jA}}{B+jC}}\right| = \frac{|e^{-jA}|}{|B+jC|} = \frac{1}{\sqrt{B^2 + C^2}} $$

Your solution is wrong because you did not calculated the absolute value. Your calculation is to get a real denominator, not the absolute value.

Answer 3

You can compute in two ways:

$$\left|\frac1{a+ib}\right|=\frac1{\left|a+ib\right|}=\frac1{\sqrt{a^2+b^2}}$$

or

$$\left|\frac1{a+ib}\right|=\left|\frac{a-ib}{a^2+b^2}\right|=\frac{\left|a-ib\right|}{a^2+b^2}=\frac{\sqrt{a^2+b^2}}{a^2+b^2}=\frac1{\sqrt{a^2+b^2}}.$$