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Partial Fractions but in Complex Analysis

  • Thread starter Fellowroot
  • Start date
  • #1
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Homework Statement



Use partial fractions to rewrite:

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



Homework Equations


none


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.
 

Answers and Replies

  • #2
2,544
2
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
HallsofIvy
Science Advisor
Homework Helper
41,805
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Homework Statement



Use partial fractions to rewrite:

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



Homework Equations


none


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

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

Your partial fractions should be
[tex]\frac{2z}{z^2+ 3}= \frac{A}{z+ i\sqrt{3}}+ \frac{B}{z- i\sqrt{3}}[/tex]
 
  • #4
92
0
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
 

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