e^i(pi) + 1 = 0

 

if we modify it like so

 

e^(i(pi)) = -1

e^(i(pi))^2 = 1

i^2(pi)^2 = ln1 = 0

 

so -pi^2 = 0

 

which is obviously wrong... so why?

Well, for a start, you're taking the logarithm of an imaginary number, which is a dodgy proposition at best.

i'm taking the log of 1, not the imaginary number

the imaginary number is just the exponent which can then be made the coefficient of ln e or 1

im only learning high school math, so there may be something that's missing (or IS rather than may)

well, the first flaw lies in you taking squares.

 

you have written[math]e^{(i\pi)^{2}}[/math] , when it should be

[math](e^{i\pi})^2[/math]

but then, you might say

 

[math](e^{i\pi})^{2}=1[/math]

[math](e^{2i\pi})=1[/math]

[math]2i\pi = \ln(1) = 0[/math] and how does that work?

 

its just like dave said, taking logarithms of complex numbers is a dodgy proposition. Lets just say it is not always true that [math]\ln{e^{z}}=z[/math] if you are working with complex variables.

 

[edit] why isn't the latex showing up?[/edit]

http://mathforum.org/dr.math/faq/faq.euler.equation.html

 

try this link. the second proof of euler's equation seems to be the one you are looking for. It says further on down to just replace x with pi to prove your equation. I can't follow it that well but i hope it makes more sense to you!

why isn't the latex showing up?

 

There's a problem with the server. blike should have it fixed either today or tomorrow.

um, i ended up with [math]\pii=\pii[/math].

we all know that lne^x=x, right. so, the left side is pi*i. i know that logs of negative numbers are imaginary, so i put ln-1 in my calc and got 3.141592654i which means pi*i. so, it does work.