I'm not exactly sure what you are asking...could you please explain more?

Ah ok. Well i don't quite know still but I'll look it up. Thanks though.

Since we're talking about math, do you mind showing me your proof? I'd like to see it personally, as I'm always interested to learn something new about math.

......Shat.
I kind of made an assumption that's not quite valid
Here it is though, if you want to see it



Ok, so Euler's Formula states

e^(xi)=cos x + i sinx  (or cis x)

Step 1: prove for x=0

e^0i= cos0 + i sin0
1=1+0i
1=1

Step 2 Assume true for x=k

e^(ki) = cis k

Step 3 Prove for x=k+1

e^(k+1)i=cis(k+1)       
e^(ki+i)=                         //distribute i
(e^ki)(e^i)=                     // exponent property
(cis k)(cis 1)=                    // here's where i mess up. In my original proof, I did induction the standard way, except it's really hard to prove e^i=cis 1,       so I just kind of assumed that it was true and used it here to sub it in as e^i=cis 1
(cos k +i sin k)(cos 1 + i sin 1)=   //expand
cos k cos 1 + i sin 1 cos k + i sin k cos 1 + -1 sin k sin 1   //multiply out
cos k cos 1 -sin k sin 1 + i (sin 1 cos k + sin k cos 1) =     //re order terms
cos (k+1) +i sin (k+1)                                                   //recognize cos and sin sum formulas
cis (k+1) =cis (k+1)

Subbing in pi  gives e^ (pi i) = cos pi + i sin pi
                                       =   -1    +   0i
                                       = -1

I always do this, I get an idea, I get lazy, skip a step, and then realize that it threatens the whole integrity of the proof >.<