MHB Partial fraction decomposition

[lucad93]

Hello everybody! I have to decompose to simple fractions the following function: $$V(z)=\frac{z^2-4z+4}{(z-3)(z-1)^2}$$. I know I can see the function as: $$V(z)=\frac{A}{z-3}+\frac{B}{(z-1)^2}+\frac{C}{z-1}$$, 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: $$A: Res(f, 3) = 1$$, that's correct; $$B=Res(f,1) = \lim_{{z}\to{1}}\left[\d{}{x}z^2-4z+4\right]=-2$$, that's correct. Finally I have to calculate the $$Res(f,1)$$, 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

 

[zzephod]

Hello everybody! I have to decompose to simple fractions the following function: $$V(z)=\frac{z^2-4z+4}{(z-3)(z-1)^2}$$. I know I can see the function as: $$V(z)=\frac{A}{z-3}+\frac{B}{(z-1)^2}+\frac{C}{z-1}$$, 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: $$A: Res(f, 3) = 1$$, that's correct; $$B=Res(f,1) = \lim_{{z}\to{1}}\left[\d{}{x}z^2-4z+4\right]=-2$$, that's correct. Finally I have to calculate the $$Res(f,1)$$, 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

Put the partial fraction back over a common denominator:
$$
V(z)=\frac{A}{z-3}+\frac{B}{(z-1)^2}+\frac{C}{z-1}=\frac{A(z-1)^2+[B+C(z-1)](z-3)}{(z-3)(z-1)^2}
$$

The coefficient of $z^2$ in the numerator is $A+C=1 $

.

 

[BestMethod]

(z^2-4z+4)/(z-3)(z-1)^2 = 1/(z-3) - 2/(z-1)^2 + C/(z-1)

Now,plugging any number into z,say 0;

-4/3 = -1/3 -2 -C .Then,

C=-1