e^(pi * i) + 1 = 0

Why? I don't understand this at all ><

e^(pi * i) = -1

-e^(pi * i) = 1

-e^(pi * i) + 1 = 0

>>4
-e^(pi * i) + 1 = 2

e^(pi * i) - e^(pi * i) = 0

e^(pi * i) + 2 = 1

do an analytic continuation from exp x to all complex values; the same with cos x and sin x; then is easy to verify(by power series) that (1) e^(iz) = cos z + isen z; doing z = pi u have the result.
You can think that (1) is heuristic based that both classes of function (exp x  and cos x + isen x) are given by the same functional equation viz:
f( x + y ) = f( x )*f( y );
sorry for my possible lack of analysis :)

for a moar geometric reason you can think also that arg( z / w ) = arg( z ) - arg( w ) like the logarithm function in the respective intervale of angles; then you can reach by heuristic and up to constants

continuing what >>8 meant
is that

even if you don't understand why e^(z*i) = cos(z) + i*sen(z)
you can know that e^(pi*i) = cos(pi) + i*sen(pi)
since cos(pi) = -1 and sen(pi) = 0
we have that e^(pi*i) = -1
then by summing 1 on both sides we have e^(pi*i) + 1 = 0

Another way to see it is to differentiate (cos(z)+i*sin(z))/e^(i*z), you get zero, which means it is a constant function, and since it's equal to 1 at z=0, it must =1 for all z, so you get the Euler formula above.

e^(pi * i) + 1 = 0

most beautiful formula

>>12
fap fap fap

>>12
is philosopher rather than mather, I guess

>>14
Oh come on. It's elegant as hell.

>>15 is in AP Calc BC so he is an EXPERT MATHEMATICIAN.

>>16
He voices the same opinion as my crazy math prof.

>>17
he's one of those "i had an epiphany when i learned ____" faggots.

a really, really bad proof (but i don't have the time for anything more fancy):
z=cos(x)+i*sin(x)
dz/dx=-sin(x)+i*cos(x)=i*(cos(x)+i*sin(x))=i*z
dz/z=i*dx
ln(z)=ix + c
z=C*exp(ix)

bad one i know

>>19
Just because a proof is quick and easy doesn't mean it's bad if it conforms to the guidelines and works. Hmmm...

http://blog.plover.com/anniversary/Euler.html

>>19
It is bad because you have not defined derivatives for imaginary numbers. Second you have assumed that d(ln(z))/dx is equal to (dz/dx)/z which you have not yet proven for imaginary numbers. Even more problematic is the fact that you have used the imaginary logarithm which you have not even defined. You could, however, define it to be the imaginary function whose derivative is 1/z for any imaginary number, which is a complex extension of the real logarithm.

>>22

Thanks for the information (>>19 here)... the reason why I would call it bad is simply because it relies on the differentiation rules for sine and cosine (which can be shown well with Euler's formula)
I found a sheet with a better proof today while cleaning up my room:
cos(x)+i*sin(x)=
cos(n*x/n)+i*sin(n*x/n)=
( cos(x/n)+i*sin(x/n) )^n= (de Moivre's formula, can be proven with complete induction)
( 1 + ( (cos(x/n) - 1)*n + i*sin(x/n)*n )/n )^n = (basically +1-1)
and now taking the limit (i'm lacking skills to argue now WHY i'm allowed to take the limit within the bracket but it gives the right result, so...)
lim((cos(x/n)-1)->0
lim(sin(x/n)*n)->x
thus:
(1+ix/n)^n which converges to exp(ix) when n->oo

>>22

an imaginary number you can treat like a constant when differentiating, right?

There's nothing to prove, it's a consequence of how we define complex exponentials.

>>22
The imaginary logarithm is just like the normal logarithm but more fun.
Since we already know r*exp(ix) is a way of expressing all complex numbers, with r as the absolute value and x as the argument,
ln(z)=ln(r*exp(ix))=ln(r)+ln(exp(ix))=ln(r)+ix, however it is multivalued since the argument goes round in circles, repeating every 2pi, so ln(z)=ln(r)+i(x+2kpi) where k is an integer

cat