$e^{\ln(-2)} = -2$ but $\ln(-2) = \ln 2+i\pi$. How does this work? I'm messing with exponential growth functions.  I noticed that I can write 
$y(t)=y(0)\alpha^t$ as 
$y(t)=y(0)e^{\ln(\alpha)t}$
(and then I can go ahead and replace $\ln(\alpha)$ with $\lambda$.)
But how do I handle when the $\alpha$ in $\ln(\alpha)$ is negative?  WolframAlpha simply evaluates 
$e^{\ln(-2)}$ as $-2$
but how do I get avoid the whole negative number issue?
WolframAlpha also calls log the natural logarithm, which is confusing. :/

With Euler's identity, $e^{i\theta}=\cos \theta+\sin\theta$. This is just a nice identity, proven to work for complex numbers.
Any complex number can be written then as :$$z=re^{i\theta}$$
where $r$ is its modulus.
Now take $\ln$ of that:
$$\ln z=\ln (r e^{i\theta})=\ln r+\ln e^{i\theta}=\ln r+i\theta $$
Now, because $e^{i\theta}=e^{i(\theta+2k\pi)}$, we have in general,
$$\ln z=\ln r+i(\theta+2k\pi)$$
