Mar 4, 2012 Thread starter #1 D dwsmith Well-known member Feb 1, 2012 1,673 $$ \frac{1}{z^2(1-z)} = \frac{A}{z^2}+\frac{B}{1-z} $$ I can't figure out how to decompose this fraction.
$$ \frac{1}{z^2(1-z)} = \frac{A}{z^2}+\frac{B}{1-z} $$ I can't figure out how to decompose this fraction.
Mar 4, 2012 Moderator #2 Chris L T521 Well-known member Staff member Jan 26, 2012 995 dwsmith said: $$ \frac{1}{z^2(1-z)} = \frac{A}{z^2}+\frac{B}{1-z} $$ I can't figure out how to decompose this fraction. Click to expand... You're missing a term. The decomposition should be of the form \[ \frac{1}{z^2(1-z)} = \frac{A}{z}+\frac{B}{z^2}+\frac{C}{1-z} \] Is it also to be assumed that $A,B,C\in\mathbb{C}$?
dwsmith said: $$ \frac{1}{z^2(1-z)} = \frac{A}{z^2}+\frac{B}{1-z} $$ I can't figure out how to decompose this fraction. Click to expand... You're missing a term. The decomposition should be of the form \[ \frac{1}{z^2(1-z)} = \frac{A}{z}+\frac{B}{z^2}+\frac{C}{1-z} \] Is it also to be assumed that $A,B,C\in\mathbb{C}$?