# Complex number in Denominator

1. Aug 26, 2010

### patm95

1. The problem statement, all variables and given/known data
Express (see attachment) with a real denominator

2. Relevant equations

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

3. 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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2. Aug 26, 2010

### Mentallic

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.

3. Aug 26, 2010

### patm95

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......

4. Aug 26, 2010

### vela

Staff Emeritus
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.

5. Aug 26, 2010

### Pengwuino

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.

6. Aug 27, 2010

### HallsofIvy

Staff Emeritus
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.)

Last edited: Sep 1, 2010
7. Sep 1, 2010

### patm95

Thanks very much for all the help. I did get this right thanks to all your help!