# Partial Fractions Complex Decomposition

• brianhawaiian
In summary, the conversation discusses how to factor and solve the equation 1/(z+1)(z2 + 2z + 2) using partial fractions. The quadratic formula is mentioned as a method to find the roots, but it is noted that the equation cannot be simplified further. The conversation also mentions the possibility of using a real decomposition or computing the residues at the poles to obtain the partial fraction decomposition.
brianhawaiian

## Homework Statement

Sorry I don't have equation editor working

1/(z+1)(z2 + 2z + 2)

## 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.

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

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?

brianhawaiian said:
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.

Dick said:
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

brianhawaiian said:
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.

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

## 1. What are partial fractions complex decomposition?

Partial fractions complex decomposition is a method used to simplify complex rational expressions into more manageable expressions. It involves breaking down a fraction with a polynomial in the numerator and denominator into smaller, simpler fractions.

## 2. Why is partial fractions complex decomposition useful?

Partial fractions complex decomposition is useful because it allows for easier integration of complex rational expressions, which is helpful in many areas of science, such as physics and engineering. It also provides a way to solve equations involving rational expressions.

## 3. How do you perform partial fractions complex decomposition?

To perform partial fractions complex decomposition, you must first factor the numerator and denominator of the rational expression. Then, set up an equation with unknown coefficients for each distinct factor in the denominator. Solve for the coefficients by equating the original expression to the sum of the partial fractions. Finally, combine the fractions and simplify.

## 4. Can partial fractions complex decomposition be used on any rational expression?

No, partial fractions complex decomposition can only be used on proper rational expressions, meaning the degree of the numerator is less than the degree of the denominator. If the degrees are equal, the expression must first be divided to make it proper.

## 5. Are there any limitations to using partial fractions complex decomposition?

Yes, there are limitations to using partial fractions complex decomposition. It can only be used on rational expressions that can be factored, and it does not work for expressions with complex roots. Additionally, the process can become very tedious and time-consuming for expressions with higher degrees or many distinct factors in the denominator.

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