Complex Numbers and Euler's EquationDate: 11/10/96 at 07:42:47 From: jingde Subject: Calculus Question on Complex Numbers Dear Dr Anthony, I have a question relating to complex numbers. This one appears to be really tough because, as to date, four senior high school maths teachers couldn't solve it. It is a question from a popular calculus textbook used here in WA: If x = Cos A + iSin A and y = Cos B + iSin B, show that: (x+y)(xy - 1) Sin A + Sin B ------------- = ------------- (x-y)(xy + 1) Sin A - Sin B Regards, Jingde Chu Date: 11/10/96 at 13:23:37 From: Doctor Anthony Subject: Re: Calculus Question on Complex Numbers To solve this problem, you need to use Euler's equation, which states that e^(ix) = Cos x + iSin x. With a little bit of manipulation, you can make the substitution: Sin A = [e^(iA) - e^(-iA)]/(2i) and Sin B = [e^(iB) - e^(-iB)]/(2i) To prove that Sin A and Sin B really do equal these expressions, substitute e^(iA) = Cos A + iSin A and e^(-iA) = Cos A - iSin A into the expression for Sin A. You see that this gives: [(Cos A + iSin A) - (Cos A - iSin A)]/2i = (Cos A - Cos A + iSin A + iSin A)/2i = 2iSin A/2i = Sin A So it really works! The same is true, of course, for Sin B. If you want to know more about Euler's equation, look at this web site: http://mathforum.org/dr.math/problems/graf.4.7.97.html Now, on with proving the equality. Using the above substitutions, the right hand side becomes: [e^(iA) - e^(-iA) + e^(iB) - e^(-iB)] ------------------------------------- [e^(iA) - e^(-iA) - e^(iB) + e^(-iB)] [x - 1/x + y - 1/y] = ------------------- [x - 1/x - y + 1/y] [x + y - (1/x + 1/y)] = ---------------------- [x - y - (1/x - 1/y)] xy(x+y) - (x+y) = --------------- xy(x-y) + (x-y) (x+y)(xy -1) = ------------ (x-y)(xy + 1) I hope that this is clear! -Doctors Anthony and Rachel, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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