Partial Fractions Complex Decomposition

[brianhawaiian]

Homework Statement

Sorry I don't have equation editor working

1/(z+1)(z² + 2z + 2)

Homework Equations

The Attempt at a Solution

(z² + 2z + 2)

z² + 1) can be factor as (z - i)(z + i) However, I'm having trouble seeing the pattern on what (z² + 2z + 2) would become, I tried guessing and checking but haven't found anything that works... I can't find a value for z that would make that equation 0.

 

[Dick]

You don't have to guess the roots. You could just use the quadratic equation to find them.

 

[brianhawaiian]

haha, okay, couldn't I just do

1/(z+1)(z² + 2z + 2)

then

1/(z+1)(z² + 2z + 2) = A/(z² + 2z + 2) + B/(z+1)

Then 1 = A(z+1) + B(z² + 2z + 2)

First let z = -1 solve for B, so I found B to be 1. Knowing that couldn't I let z = anything, and solve for A?

 

[Dick]

haha, okay, couldn't I just do

1/(z+1)(z² + 2z + 2)

then

1/(z+1)(z² + 2z + 2) = A/(z² + 2z + 2) + B/(z+1)

Then 1 = A(z+1) + B(z² + 2z + 2)

First let z = -1 solve for B, so I found B to be 1. Knowing that couldn't I let z = anything, and solve for A?

You could try. But you'd find what you get for A is dependent on what you pick for z. There's no A and B that will work for all z. (z^2+2z+2) is the product of two linear factors. You have to split it up completely to make partial fractions work.

 

[brianhawaiian]

You could try. But you'd find what you get for A is dependent on what you pick for z. There's no A and B that will work for all z. (z^2+2z+2) is the product of two linear factors. You have to split it up completely to make partial fractions work.

Got it, thanks a million

 

[Dick]

Got it, thanks a million

Or you could try writing it as (A*z+B)/(z^2+2z+2)+C/(z+1) if you want a real decomposition. Forgot to mention that possibility.

 

[Count Iblis]

You can also obtain the partial fraction decomposition by computing the residues at the poles.