# Homework Help: Partial Fractions but in Complex Analysis

1. Nov 27, 2011

### Fellowroot

1. The problem statement, all variables and given/known data

Use partial fractions to rewrite:

(2z)/(z^2+3)

2. Relevant equations
none

3. The attempt at a solution

I did this:

(2z)/(z^2+3) = (Az+B)/(z^2+3)

2z = Az +B

A = 2, B = 0......problem is that it just recreates the original

Here is their example in the book:

1/(z^2+1) = 1/(2i(z-i)) - 1/(2i(z+i))

I don't understand how they came to their conclusion in their example.

Any help understanding their example or my question is appreciated.

2. Nov 27, 2011

### cragar

factor 1/(z^2+1) as 1/(z+i)(z-i)
then 1/(z^2+1)= A/(z+i)+ B/(z-i)
now solve for A and B.... what do you get .
and then you can do the same trick on your problem

3. Nov 27, 2011

### HallsofIvy

Of course it does. You wrote it in exactly the same form. Why would you expect anything else?

You titled this "Partical fractions but in Complex Analyis"- so use complex numbers:
$z^2+ 3= (z+ i\sqrt{3})(z- i\sqrt{3})$.

$$\frac{2z}{z^2+ 3}= \frac{A}{z+ i\sqrt{3}}+ \frac{B}{z- i\sqrt{3}}$$

4. Nov 27, 2011

### Fellowroot

Thanks guys, oh my gosh

I actually did their example and set it up all correctly

I got to the end with A+B=0, A-B=0 and without solving it, it just looked like a contradiction and I didnt realize that B=-1.

thanks for pointing that out for me, i was like 99% there and got stuck. i hate that