# Partial Fraction From Hell!

1. Aug 30, 2007

### WolfOfTheSteps

1. The problem statement, all variables and given/known data

I know how to use the method of partial fractions in most circumstances, but I'm working on a problem that has gotten the best of me. How do I get from the left side of the following identity to the right side????

$$\frac{-2-2\omega^2}{-\omega^2+\sqrt{2}i\omega+1} \ = \ \ 2 \ + \ \frac{-\sqrt{2}-2\sqrt{2}i}{i\omega - \frac{-\sqrt{2}+i\sqrt{2}}{2}} \ + \ \frac{-\sqrt{2}-2\sqrt{2}i}{i\omega - \frac{-\sqrt{2}-i\sqrt{2}}{2}}$$​

3. The attempt at a solution

I was able to factor the denominator and write the following equation:

$$A\left(\frac{\sqrt{2}+i\sqrt{2}}{2} +i\omega\right) \ + \ B\left(\frac{\sqrt{2}-i\sqrt{2}}{2} +i\omega\right) \ = \ -2-2\omega^2$$​

but couldn't get much further because A and B don't have $\omega^2$ multiples to match up with the $-2\omega^2$ on the right side of the equation.

How do I handle this monster?

Note: $i$ is the imaginary unit.

Thanks!

2. Aug 30, 2007

### HallsofIvy

Staff Emeritus
I would like you to show how you arrived at what you give since that is not what I get.

The first thing I did was multiply both numerator and denominator by -1 to get
$$\frac{2+2\omega^2}{\omega^2-\sqrt{2}i\omega-1}$$
Then the denominator factors as
$$\omega^2- \sqrt{2}i\omega- 1= (\omega-\frac{\sqrt{2}}{2}-i\frac{\sqrt{2}}{2})(\omega+\frac{\sqrt{2}}{2}-i\frac{\sqrt{2}}{2})$$
Putting those factors into the denominators and multiplying through by them gives
$$A(\omega+\frac{\sqrt{2}}{2}-i\frac{\sqrt{2}}{2})+ B(\omega-\frac{\sqrt{2}}{2}-i\frac{\sqrt{2}}{2})= 2+ 2\omega^2$$

Now set
$$\omega= \frac{\sqrt{2}}{2}+ i\frac{\sqrt{2}}{2}$$
and
$$\omega= \frac{\sqrt{2}}{2}- i\frac{\sqrt{2}}{2}$$
and it should be easy.

3. Aug 30, 2007

### WolfOfTheSteps

Your denominator factors are not in the form I need them to be. I didn't check it, but I assume your work is correct. The problem is I need the denominator factors to be in the $(a +i\omega)$ (where $a$ is complex or real) form to take advantage of a known Fourier transform.

Here is how I arrived at my equation:

Let

$$(A+i\omega)(B+i\omega) = 1 + \sqrt{2}i\omega-\omega^2$$

Then we want A and B to satisfy the following conditions:

$$A+B = \sqrt{2}$$
$$AB = 1$$

Solving that system we get:

$$A= \frac{\sqrt{2}}{2}(1-i)$$
$$B= \frac{\sqrt{2}}{2}(1+i)$$

Hence,

$$1+\sqrt{2}i\omega - \omega^2 = \left(\frac{\sqrt{2}}{2}(1-i)+i\omega\right) \left(\frac{\sqrt{2}}{2}(1+i)+i\omega\right)$$

And from that I was able to get the equation I gave before:

$$A\left(\frac{\sqrt{2}+i\sqrt{2}}{2} +i\omega\right) \ + \ B\left(\frac{\sqrt{2}-i\sqrt{2}}{2} +i\omega\right) \ = \ -2-2\omega^2$$

And that's where I was stuck.

Last edited: Aug 30, 2007
4. Aug 31, 2007

### WolfOfTheSteps

HallsofIvy,

I got it now. I found out that I have to first do long division to get the numerator to be 1 degree less than the denominator, and then use the method of partial fractions.