- #1
lucad93
- 4
- 0
Hello everybody! I have to decompose to simple fractions the following function: \(\displaystyle V(z)=\frac{z^2-4z+4}{(z-3)(z-1)^2}\). I know I can see the function as: \(\displaystyle V(z)=\frac{A}{z-3}+\frac{B}{(z-1)^2}+\frac{C}{z-1}\), and that the terms A, B, C can be calculated respectively as the residues in 3 (single pole), 1 (double pole), and 1 (single pole). So i calculate: \(\displaystyle A: Res(f, 3) = 1\), that's correct; \(\displaystyle B=Res(f,1) = \lim_{{z}\to{1}}\left[\d{}{x}z^2-4z+4\right]=-2\), that's correct. Finally I have to calculate the \(\displaystyle Res(f,1)\), single pole. I tried in some ways: by using the single pole formula and then using the "de l'hopital"'s theorem, by calculating the decomposed function in 0 and then equaling to the function in 0, but nothing was correct. Can you tell me a correct way to proceed? Thankyou