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
 A: 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 $\sin(-t)=-\sin t$.
A: This is wrong.  In fact, both cosine and sine could be complex, but the imaginary bits cancel out.
You can find what it comes out to on Wikipedia.
A: If $z$ is a complex number and $z=u+iv$,  it does not follow that $u=Re(z)$ and $v= Im(z)$ !
Example: $2+2i$ can be written as
$$(1+i)+i(1-i).$$
A: It is an interesting exercise! The problem you're having though, is that you don't know what $\sin(ai)$ and $\cos(ai)$ are. So instead, you'll may want to try and verify that their outcomes are not complex.
Hint: Try writing the taylor expansion of $e^{-a},\cos{(ai)}$ and $\sin{(ai)}$
A: The thing is $\sin{(ai)}$ is complex and not real, so you cannot say $\cos{(ai)}$ and $\sin{(ai)}$ both equal to zero. 
