Understanding an equality: $|e^{-f(z)}|= e^{-Re(f)}$ Why is the following equality true: $$|e^{-f(z)}|= e^{-Re(f)}$$
 A: Hint:
$$|e^{z}|=|e^{x+yi}|=|\underbrace{e^x}_{>0}|\underbrace{|e^{yi}|}_{=1}=e^x$$
EDIT: there are two ways to show that $e^{iy}$ has modulus $1$. The first is to recall Euler's identity:
\begin{align}|e^{yi}|^2& =|\cos(y)+i\sin(y)|^2\\
& =(\cos(y)+i\sin(y))\cdot \overline{\cos(y)+i\sin(y)}\\& =(\cos(y)+i\sin(y))(\cos(y)-i\sin(y))\\ &=\cos^2(y)+\sin^2(y)\\ &=1,
\end{align}
where the last step is just pythagoras. Then the results follows as the modulus is a nonnegative real number.
Alternatively using directly that $\overline{e^{iy}}=e^{-iy}$ we obtain
$$|e^{yi}|^2=e^{yi}\overline{e^{iy}}=e^{yi}e^{-iy}=e^{iy-iy}=e^0=1.$$
Geometrically you can see the $e^{iy}$ as the point on the unit circle, enclosing an angle of size $y$ with the Origin and the positive $x$-Axis.
A: $$|\exp(a+bi)| = |\exp(a)\exp(bi)| = |\exp(a)||\cos b + i \sin b|$$
$$=\exp(a)\sqrt{\cos^2 b+\sin^2 b}= \exp(a).$$
A: Hint
Write $f(z)=x+yi$
$$|e^{-x-yi}|=|e^{-x}\cdot e^{-yi}|=|e^{-x}|\cdot|e^{-yi}|=e^{-x}$$
Because $e^{-x}>0$ and by Euler formula $|e^{-yi}|=1$
A: $$|e^{-f}|=|e^{-{\bf Re}f-i{\bf Im}f}|=|e^{-{\bf Re}f}||e^{-i{\bf Im}f}|=e^{-{\bf Re}f}$$
A: By splitting $f(x) = u(z) + i \cdot v(z)$ into the real and imagenary parts of $f$, on can simply see: 
$$
e^{f(z)} = e^{u(z)+ i \cdot v(z)} = e^{u(z)} \cdot e^{i \cdot v(z)}$$
Observe, that $|e^{i \cdot q}| = 1 \quad \forall q$, so it doesn't contribute to the magnitude of the complete equation. 
The minus doesn't change that. 
