Evaluate $(-1)^x$ by hand. How to calculate the value for $(-1)^x$ for any $x$ by hand.  Using Mathematica I kind of figured $(-1)^x=\cos(\pi x)+i\sin (\pi x)$. but how can I prove this.
This is my first question here. Sorry for the mistakes.
 A: One problem is that $(-1)^x$ is not well defined, since $-1=e^{ki\pi}$ for no matter which odd number $k$ you choose. You may multiply the exponent by $x$ and get, in most cases, infinitely many different values for the value of the exponential. I would be very surprised indeed if you could do this calculation by hand, and I will be delighted to be so surprised.
A: $$-1^x=e^{x\log (-1)}=e^{xi\pi+2\pi ik}=i\sin(x\pi+2\pi k)+\cos(x\pi+2\pi k),k\in \Bbb{Z}$$
You may want to use a few taylor expansion terms of cos(x) and sin(x) to find your value accurately 
A: Complex exponentiation is not simple. Most difficulties can be solved by giving a  definition for the logarithm.  Since $e^{i\pi k}=-1$ for any odd integer $k$, there are an infinite number of possibilities for $\log(-1)$. If we use $\log(-1)=\pi i$, then your formula is correct; that is,
$$
(-1)^x=e^{x\log(-1)}=e^{i\pi x}=\cos(\pi x)+i\sin(\pi x)
$$
However, we could also define $\log(-1)=-\pi i$, then
$$
(-1)^x=e^{x\log(-1)}=e^{-i\pi x}=\cos(\pi x)-i\sin(\pi x)
$$
Once you settle on a branch of the logarithm, exponentiation can be well-defined.
