exp(x) for imaginary numbers Well, I know how to get the $e^x$ function polynomial expansion, but how do I know that this is also valid for imaginary numbers, like $i\pi$?
I know that the Euler's Formula is:
$$e^{ix} = \cos(x) + i\sin(x)$$
It means that both the $\cos(x)$ and $\sin(x)$ expansions allow na imaginary number like $i\pi$ to be the input. How can I prove this? Is there some good PDF about this?
 A: Defining the complex exponential is straightforward.
What we want to do is find an analytic function $f(z)$ such that $$f(z_1+z_2) = f(z_1)f(z_2)$$ and $$f(x) = e^x$$
for all real $x$.
Therefore, we have $f(x+iy) = f(x)f(iy) = e^xf(iy)$, where the latter equality comes from the second condition.
Define $f(iy) \equiv M(y)+iN(y)$, which is just another way of saying "break $f$ into real and complex parts." Note that this means that $f(x+iy) = e^xM(y) + e^x iN(y)$. Since we want the function to be analytic, it must satisfy the Cauchy-Riemann equations:
$$\begin{align*}
M(y) &= N'(y), \\
M'(y) &= -N(y),
\end{align*}$$
hence, $$M''=-M.$$
Solving this differential equation gives us $M = a \cos y + b \sin y$ and so $N = -M' = a\sin y - b\cos y$.
Now, we use that $f(x) = e^x$ for all real $x$, which is true whenever $y = 0$. Since $M(0) = a$ and $N(0) = -b$, we use $$ e^x = f(x) = f(x+i\cdot 0) = e^xM(0)+e^x iN(0)$$
to conclude that $a=1$ and $b=0$.
Hence, $e^z = e^{x+iy} = e^x \cos y+ ie^x\sin y$.
This is why they Taylor series works.
