what?? I is -1 square rooted, and that isnt a real number.
It's actually quite simple if you understand taylor series. This is, I think, the most comprehensible proof of Euler's Formula e^(ix)=cos(x)+isin(x), which is why the thread title statement is true.
I'll summarize the proof on this page as well to perhaps keep page toggling down if anyone is interested, note though that the important part is at the bottom of the post:
The taylor series for e^x is:
e^x=1 + x + (x^2)/2! + (x^3)/3! + (x^4)/4! + ... (x^n)/n!
Now plug in ix to x to get the taylor series for e^(ix):
e^(ix)=1 + ix + ((ix)^2)/2! + ((ix)^3)/3! + ((ix)^4)/4! + ....
Apply the power to the i to bring it out:
e^(ix)=1 + ix - (x^2)/2! - i(x^3)/3! + (x^4)/4! + i(x^5)/5! - (x^6)/6! + ...
Note that if you group the terms with i together and those without i together and them factor out the i from the group with the i's in it...
e^(ix)=(1 - (x^2)/x! + (x^4)/4! - (x^6)/6! + ...) + i(x - (x^3)/3! + (x^5)/5! - (x^7)/7! + ...)
...You wind up with the red being the taylor series for cosine and blue being the taylor series for sine. Thus, you can justify replacing the series with the functions themselves, giving Euler's Formula:
e^(ix)=cos(x)+isin(x)
//=========================
So we now have the formula e^(ix)=cos(x)+isin(x).
Now, we're looking for the value of e^(ix) when x is pi, thus e^(i*pi). Well, plug in pi for x to get:
e^(i*pi)=cos(pi)+i*sin(pi)
e^(i*pi)=-1 + i*0
e^(i*pi)=-1