This question arises actually from electrical engineering but I think it is not hard-bounded to it and the math itself isn't much special anyway, I guess? So my question is:
There is a function $ u(t) = A \cdot e^{(-at)} $ and you have to apply a Fourier transformation which results in $ U(jw) = \frac{A}{a+jw} $. Up to here everything is fine and I get it. Next task is to determine the so-called amplitude response $ |U(jw)| $.
What I know is the following:
$ x_Z(jw) = | x_Z(jw) | \cdot e^{j \Phi_Z (w)} $ and $ x_N(jw) = | x_N(jw) | \cdot e^{j \Phi_N (w)} $ and together you'd have: $ x(jw) = \frac{| x_Z(jw) |}{| x_N(jw) |} = | x(jw) | \cdot e^{j \Phi (w)} $
This should be applied somehow (I guess, I'm not sure) to $ U(jw) $ above and yield:
$ x(jw) = \frac{A}{a+jw} = \frac{A}{\sqrt{a^2 + w^2} \cdot e^{jarctan \Phi (\frac{w}{a})}} $
I don't understand this lest step. How do I get from $ x(jw) = \frac{A}{a+jw} $ to $ x(jw) = \frac{A}{\sqrt{a^2 + w^2} \cdot e^{jarctan \Phi (\frac{w}{a})}} $ ?
edit: Following up on Eric's answer my question is then: How do I know what $ x $ and $ y $ are in $ U(jw) = \frac{A}{a+jw} $ ? I guess it is $ x = a $ and $ y = w $ but what about the $ A$ ?
$a$ and $w$ (possibly $\omega$ in the original?) are real numbers, so $|a + \mathrm{j}w| = \sqrt{a^2 + w^2}$ by the Pythagorean theorem. Then $\arg(a + \mathrm{j}w)$ is either $\arctan(w/a)$ or $\pi + \arctan(w/a)$ (depending on which quadrant $a + \mathrm{j}w$ is in, which is the job being performed by $\Phi$) for most $a$ and $w$. Really, what is being done is converting a complex number in Cartesian coordinates to polar coordinates. See polar complex plane for more details. (The arctangent used at the Wikipedia article is a little sneaky, to avoid having to work out quadrants. You can imagine that $\Phi$ is here to implement equivalent sneakiness.)
So we have $u(t) = A \mathrm{e}^{-at}$. The Laplace transform of $u(t)$ is $U(\mathrm{j}w) = \frac{A}{a + \mathrm{j}w}$. We want to split this into its magnitude part and angle part. The numerator is a real number so $$ A = \begin{cases}|A| \mathrm{e}^{\mathrm{j} 0} ,& A \geq 0 \\ |A| \mathrm{e}^{\mathrm{j} \pi} ,& A < 0 \end{cases} $$ With that, we're ready to work out the polar form of $U(wt)$. First we break the numerator into its polar form and do the same to the denominator. \begin{align*} U(wt) &= \frac{A}{a+wt} \\ &= \begin{cases} \frac{|A|\mathrm{e}^{\mathrm{j} 0}}{\sqrt{a^2 + w^2} \mathrm{e}^{\mathrm{j} \arctan \Phi(w/a)}} ,& A \geq 0 \\ \frac{|A|\mathrm{e}^{\mathrm{j} \pi}}{\sqrt{a^2 + w^2} \mathrm{e}^{\mathrm{j} \arctan \Phi(w/a)}} ,& A < 0 \end{cases} \\ &= \begin{cases} \frac{|A|}{\sqrt{a^2 + w^2}} \frac{\mathrm{e}^{\mathrm{j} 0}}{ \mathrm{e}^{\mathrm{j} \arctan \Phi(w/a)}} ,& A \geq 0 \\ \frac{|A|}{\sqrt{a^2 + w^2}}\frac{\mathrm{e}^{\mathrm{j} \pi}}{\mathrm{e}^{\mathrm{j} \arctan \Phi(w/a)}} ,& A < 0 \end{cases} \end{align*} From here we can see that for either sign of $A$, the magnitude of $U$ is $\frac{|A|}{\sqrt{a^2 + w^2}}$. (And the angle is something a little complicated, either $- \arctan \Phi(w/a)$ or $\pi - \arctan \Phi(w/a)$, that depends on the sign of $A$. If we already knew $A > 0$, then the case for negative $A$ is unneeded.)