How to take an absolute value or modulus of z?

Question

let's assume the :

$$z=\frac{e^{(-jc)}}{(a+jb)}$$ I would like to take the absolute value of z. I started with multiplication z with $\frac{(a-jb)}{(a-jb)}$ and got: $$abs\frac{e^{(-jc)}}{(a+jb)}=\frac{e^{(-jc)}(a-jb)}{(a^2+b^2)}$$ Then: $$\frac{e^{(-jc)}(a-jb)*e^{(jc)}}{(a^2-b^2)e^{(jc)}}=\frac{(a-jb)}{(a^2+b^2)e^{(jc)}}$$ Should I repeat the calculations or I made something wrong?

edit 1:

$(a-jb)(a+jb)=a^2+b^2$

edit 2:

$ abs(e^{(-jc)})=1$

$ abs(z)=\frac{(a-jb)}{(a^2+b^2)e^{(jc)}}$

what should i do with $(a-jb)$? Or will be it $(a^2+b^2)$?

Answer 1

Hint: If $w=re^{j\theta}$ for $r,\theta\in\Bbb R$, then $|w|=|r|$.

Answer 2

Recall that $$ \left|\frac{z_1}{z_2}\right|=\frac{|z_1|}{|z_2|} $$ If $c$ is real, then $|e^{-jc}|=|\cos c+j\sin c|=1$. So you have $$ \left|\frac{e^{-jc}}{a+jb}\right|=\frac{1}{|a+jb|}=\frac{1}{\sqrt{a^2+b^2}} $$