# Partial fraction decomposition: One quick question

1. Apr 24, 2013

### nate9228

1. The problem statement, all variables and given/known data
Give the partial fraction decomposition of 1/z4+z2

2. Relevant equations

3. The attempt at a solution
My question is about the final answer. The book gives the answer to be 1/z2+ 1/2i(z+i)- 1/2i(z-i). For my answer I keep getting a negative for both of the 1/2i coefficients, i.e my answer has a - where the book puts a +, and I can not for the life of me figure out what I am missing. My answer revolves around 1= A(z+i)(z-i)+ Bz2(z-i)+ Cz2(z+i) and then solving for the coefficients. How is B= 1/2i and not -(1/2i)?

2. Apr 24, 2013

### HallsofIvy

Staff Emeritus
First, write it correctly! What you wrote would normally be interpreted as $(1/z)^4+ z^2$ but I am sure you mean $$1/(z^4+ z^2)= 1/[z^2(z+i)(z- i)]$$. Normally, I would look for partial fractions with real coefficitents, but with tthe "book answer" you give it must be of the form
$$\frac{1}{z^2(z+ i)(z- i)}= \frac{A}{z}+ \frac{B}{z^2}+ \frac{C}{z+ i}+ \frac{D}{z- i}$$

Multiplying both sides by $$z^2(z+ i)(z- i)$$ gives $$1= Az(z- i)(z+ i)+ B(z+ i)(z- i)+ Cz^2(z- i)+ Dz^2(z+ i)$$

Taking z= 0, 1= B. Taking z= i, 1= D(-1)(2i) so that D= -1/2i= (1/2)i. Taking z= -i, 1= C(-1)(-2i)= 2iC so that C= 1/2i or -(1/2)i. I think that last is what you are asking about.

Last edited by a moderator: Apr 25, 2013