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### 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/
```
Associated Topics:
High School Imaginary/Complex Numbers

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