DenX2 Posted March 15, 2005 Share Posted March 15, 2005 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? Link to comment Share on other sites More sharing options...
Dave Posted March 15, 2005 Share Posted March 15, 2005 Well, for a start, you're taking the logarithm of an imaginary number, which is a dodgy proposition at best. Link to comment Share on other sites More sharing options...
DenX2 Posted March 15, 2005 Author Share Posted March 15, 2005 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) Link to comment Share on other sites More sharing options...
bloodhound Posted March 16, 2005 Share Posted March 16, 2005 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] Link to comment Share on other sites More sharing options...
dan19_83 Posted March 16, 2005 Share Posted March 16, 2005 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! Link to comment Share on other sites More sharing options...
Dave Posted March 16, 2005 Share Posted March 16, 2005 why isn't the latex showing up? There's a problem with the server. blike should have it fixed either today or tomorrow. Link to comment Share on other sites More sharing options...
ydoaPs Posted March 16, 2005 Share Posted March 16, 2005 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. Link to comment Share on other sites More sharing options...
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