Trigonometric functions for complex numbers I thought I had defined what cos(ai) and sin(ai) was earlier today when I did the following:

$e^{(vi)} = \cos(v) + i\sin(v)$ 
If we let $v = ai$, where a is real, we get:
$e^{(aii)} = \cos(ai) + i\sin(ai) = e^{(-a)}$
Since $e^{(-a)}$ is a real number, the $i\sin(ai)$ must be 0, and therefor $\cos(ai)$ must be $e^{(-a)}$.
So I have concluded that $\cos(ai) = e^{(-a)}$ and $\sin(ai) = 0$ 

I am pretty sure this is wrong since I have seen different answers online, but I would like to know what I did wrong.

Thanks
It's worth knowing these identities:

$$\cos it = \cosh t$$
$$\sin it = i \sinh t$$

I presume you are familiar with the hyperbolic functions $\cosh$ and $\sinh$. These are usually defined as $\cosh t \equiv \frac12(e^t+e^{-t})$, and $\sinh t \equiv \frac12(e^t-e^{-t})$.

These functions have properties which parallel those of the circular functions.

So, it is indeed true that $\cos it$ is a real number when $t$ is real.

Addendum: An easy way to remember this is that it is completely analogous to how the sign is handled with circular functions: $\cos(-t)=\cos t$ and $\sin(-t)=-\sin t$.
The thing is $\sin{(ai)}$ is complex and not real, so you cannot say $\cos{(ai)}$ and $\sin{(ai)}$ both equal to zero.