How can complex numbers be expressed with a real denominator?

AI Thread Summary
To express complex numbers with a real denominator, one can multiply the numerator and denominator by the complex conjugate of the denominator. This process, known as rationalizing the denominator, involves converting the denominator into the form a + ib and then applying its conjugate a - ib. The discussion highlights that if r and δ are real, the conjugate can be simplified by merely changing the sign of the imaginary unit. The final expression achieved is a fraction where the denominator is guaranteed to be real. This method effectively resolves the issue of complex denominators in mathematical expressions.
patm95
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Homework Statement


Express (see attachment) with a real denominator


Homework Equations



Not sure if there is really relevant equations to use here.

The Attempt at a Solution



First I multiply the top and bottom by the exponential. That gives me e^ix/(e^ix-r^2). I think this is a good first step, but I am unsure where to go from here.
 

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To make the denominator real (called rationalizing the denominator) you need to convert it into the form a+ib and then multiply by its complex conjugate a-ib. Can you turn it into that form first? It will be easier if you convert eix into its polar form expression.
 
If I convert the denominator as given in the problem I end up with: i r^2 sin(x)+r^2 (-cos(x))+1 I don't see how I can make a conjugate with 3 variables here? Am I missing something? I have went over this several times in attempt to make sure I hadn't made a mistake...
 
I take it r and δ are real. If that's the case, it's a bit simpler to simply replace i with -i to form the conjugate of the denominator, rather than converting to rectangular form. Use the fact that z+\bar{z} = 2 \textrm{Re}(z) to simplify the cross term.
 
Whenever you have a problem of the form \frac{a}{b} where 'a' and 'b' are in general, complex, you can immediately multiply top and bottom by the conjugate of b.

\frac{a}{b} \frac{b^*}{b^*}

and the result in the denominator is b*b = |b|, which is always real.

If r, \delta are real, the only change when you conjugate is changing the sign on the i's.
 
You don't have to write the complex number as a+ bi. Just multiply both numerator and denominator by the complex conjugate, 1- r^2e^{i\delta}. You get
\frac{1- r^2e^{i\delta}}{1- r^2(e^{i\delta}+ e^{-i\delta})+ r^4}
= \frac{1- r^2e^{i\delta}}{1+ r^4- 2r^2 cos(\delta)}

(I see now that Vela had said essentially the same thing.)
 
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Thanks very much for all the help. I did get this right thanks to all your help!
 
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