Assume $r>0$. Instead of looking at $r\exp(\mathrm i\theta)$ replace $r$ by $\log r$ and you get $z=\exp(r+\mathrm i\theta)$.
Also $\exp(a+\mathrm i b) = \exp(a)\exp(\mathrm i b)$. This way you have an identity that simply expresses each complex number by $\exp(z)$ for some complex $z$.
Now in your case: If $\theta$ is complex, let’s say, $\theta = a+b\mathrm i$. Then $\mathrm i\theta= \mathrm i a - b$. So
$$ r\exp(\mathrm i\theta) = r\exp(-b)\exp(\mathrm i a) $$
Thus there does not really exist an imaginary angle. $\exp$ is the homomorphism that maps $\mathbb C$ to $\mathbb C^\ast$. Then the angle is basically the imaginary part of the inverse.