Partial Fractions

Dari

Lifer
Oct 25, 2002
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f(z) = z/(z^2 +z +1)^2

Everytime I work this out using the usual method, I always come back to the same formula.

The book method doesn't seem to work, unless I do some type of factorization that I'm not aware of. Anybody want to take a stab at this? It's a good way to let time fly.
 

DrPizza

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The denominators of the partial fractions would be your quadratic, and your quadratic squared. Hmmm... how do I say this?

If you had something over the quadratic, then when you added the terms together, by getting a common denominator, you would have a z^2 term in the first fraction. Since you don't have a z^2 term in the fraction you're trying to break up, there isn't going to be something over the (z^2 + z + 1) denominator. Thus, there's only going to be one fraction.


Here's a simpler example. Break up 4/x^2. A/x + B/x^2. When you get a common denominator, the first term is going to be Ax/x^2. There's nothing in the second term to cancel out that Ax, so the numerator of the first term will be of a higher degree than what you originally started with for a numerator.


There's always the chance I'm wrong, but I simply don't think it breaks up. (And I went through the long way to see if I missed something.)
 

Dari

Lifer
Oct 25, 2002
17,133
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Originally posted by: DrPizza
The denominators of the partial fractions would be your quadratic, and your quadratic squared. Hmmm... how do I say this?

If you had something over the quadratic, then when you added the terms together, by getting a common denominator, you would have a z^2 term in the first fraction. Since you don't have a z^2 term in the fraction you're trying to break up, there isn't going to be something over the (z^2 + z + 1) denominator. Thus, there's only going to be one fraction.


Here's a simpler example. Break up 4/x^2. A/x + B/x^2. When you get a common denominator, the first term is going to be Ax/x^2. There's nothing in the second term to cancel out that Ax, so the numerator of the first term will be of a higher degree than what you originally started with for a numerator.


There's always the chance I'm wrong, but I simply don't think it breaks up. (And I went through the long way to see if I missed something.)

Thanks. I understand what you mean. But it must be different with complex variables. I hate problems I can't solve because it means I can't do anything else until it's solved:).
 

Dari

Lifer
Oct 25, 2002
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Originally posted by: FleshLight
complete square?

Right. That's what I was trying to do. I'm going to try to take another stab at it. Have any hints?
 

DrPizza

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Mar 5, 2001
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Ick. I missed the complex variables part. Ick ick ick... I've completely forgotten how to do this. I tried to "reinvent" the process, but fell flat on my face, or rather, was staring at a mess that I didn't want to finish.
 

DrPizza

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dammit... I'm just like you. I can't put a problem down til I finish it. I hate this problem. I'm making lots of progress after a few "ohh, duhhh's."

crud. I was checking my result and have at least 1 sign error. Might be carelessness, but I've gotta do the dishes before my wife gets home. Maybe I'll play with this problem more later. Too much stuff I've forgotten how to do in the last 10 years. (I know there's an easier method than the one I'm using.)
 

Dari

Lifer
Oct 25, 2002
17,133
38
91
Originally posted by: DrPizza
dammit... I'm just like you. I can't put a problem down til I finish it. I hate this problem. I'm making lots of progress after a few "ohh, duhhh's."

crud. I was checking my result and have at least 1 sign error. Might be carelessness, but I've gotta do the dishes before my wife gets home. Maybe I'll play with this problem more later. Too much stuff I've forgotten how to do in the last 10 years. (I know there's an easier method than the one I'm using.)

:) Thanks a lot.