Conquering the Monstrous Partial Fraction: Solving Complex Equations

[WolfOfTheSteps]

Homework Statement

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?

<br /> \frac{-2-2\omega^2}{-\omega^2+\sqrt{2}i\omega+1}<br /> \ = \<br /> \ 2 \ + \<br /> \frac{-\sqrt{2}-2\sqrt{2}i}{i\omega - <br /> \frac{-\sqrt{2}+i\sqrt{2}}{2}} \ + \<br /> \frac{-\sqrt{2}-2\sqrt{2}i}{i\omega - <br /> \frac{-\sqrt{2}-i\sqrt{2}}{2}}<br />​

The Attempt at a Solution

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

<br /> A\left(\frac{\sqrt{2}+i\sqrt{2}}{2} +i\omega\right) \ + \<br /> B\left(\frac{\sqrt{2}-i\sqrt{2}}{2} +i\omega\right) \ = \<br /> -2-2\omega^2<br />​

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!

 

[HallsofIvy]

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.

 

[WolfOfTheSteps]

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

Thanks for your help!

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)<br /> \left(\frac{\sqrt{2}}{2}(1+i)+i\omega\right)

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

<br /> A\left(\frac{\sqrt{2}+i\sqrt{2}}{2} +i\omega\right) \ + \<br /> B\left(\frac{\sqrt{2}-i\sqrt{2}}{2} +i\omega\right) \ = \<br /> -2-2\omega^2<br />

And that's where I was stuck.

 

[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.

Thanks for your help!

 

[HallsofIvy]

Dang, didn't even occur to me!