Complex Numbers and Euler's Equation

Date: 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/