Let: $$z=a+b i$$
Prove that: $$|z|=\Re(z \exp (-i \Im(\log (z))))$$
As a starting point Mathematica tells me that: $$\Re(z \exp (-i \Im(\log (z))))=a \cos (\Im(\log (a+b i)))+b \sin (\Im(\log (a+b i)))$$
And I know that: $$|z|=\sqrt {a^2+b^2}$$ by the definition in Wikipedia and from my mathematical handbook.
How do I proceed from there?
The formula only makes sense if $\lvert z \rvert > 0$, since otherwise the log isn't defined. Let $a=r\cos{\theta}$, $b=r\sin{\theta}$, $r > 0$, $0 \leq \theta < 2\pi $. Then by Euler's formula, $z=re^{i\theta}$ and $$ \log{z} = \log{r} + i(\theta+2n\pi) $$ for some integer $n$. So the imaginary part is $\theta+2n\pi$. Then $$ \exp{(-i\Im(\log{z}))} = e^{-i\theta - 2n\pi i} = e^{-i\theta} $$ But then $$ z\exp{(-i\Im(\log{z}))} = z e^{-i\theta} = re^{i\theta-i\theta} = r = \lvert z \rvert, $$ and so in fact the last $\Re$ is unnecessary.
To do it only using $a$ and $b$ is a pain, firstly because finding the imaginary part of the logarithm's unpleasant, and then finding $e^{-i \Im(\log{z}) }$ is even worse. Much better to work in polar form.