On Fri, 28 Apr 2006, Steve Gray wrote:
Let x=0, y=Pi. Then e^(x+iy)=e^0(cos(Pi) + i sin(Pi)) = 1(-1+0) = -1.
This is my point. If e^(x+iy)=e^x (cos y + i sin y) is the definition WE have chosen, then e^(i pi)=-1 is a simple consequence thereof. However, if we say that e^(x+iy)=e^x (cos y + i sin y) is the ONLY definition such that the "new" function e^(x+iy) satisfies this and this, then indeed, e^(i pi) =-1 is a remarkable fact. Emeric
----- Original Message ----- From: "Emeric Deutsch" <deutsch@duke.poly.edu> To: "Steve Gray" <stevebg@adelphia.net> Cc: "math-fun" <math-fun@mailman.xmission.com> Sent: Friday, April 28, 2006 3:28 PM Subject: Re: [math-fun] Re: favorite theorem
On Fri, 28 Apr 2006, Steve Gray wrote:
... On these scores, e^(i pi)+1=0 is unbeatable.
I am sure that I am missing here something. What is the definition of e^(x+iy) ? Isn't it e^(x+iy)=e^x (cos y + i sin y) ?
Emeric
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