91Ó°ÊÓ

Given the function \(f(z)=\frac{4 z^{2}-2 z+8}{z^{3}-z^{2}+4 z-4}\), we want to divide it into partial fractions. Firstly, we need to factor the denominator. We have a hint that one of the roots is \(z=+1\). When \(z=1\), the denominator is equal to zero: \(1^{3}-1^{2}+4(1)-4=0\). Therefore, \((z-1)\) is a factor of the denominator. We can perform polynomial long division or synthetic division to find the other factor. Dividing \((z^{3}-z^{2}+4 z-4)\) by \((z-1)\), we find that the remaining factor is \((z^{2}+4)\). Thus, the denominator becomes \((z-1)(z^{2}+4)\). The partial fractions decomposition will have the form: $$ f(z)=\frac{A}{z-1} + \frac{Bz+C}{z^{2}+4}, $$ where \(A\), \(B\), and \(C\) are constants to be determined.

The poles of \(f(z)\) are the values of \(z\) that make the denominator zero. Since we have factorized the denominator as \((z-1)(z^{2}+4)\), the poles are \(z=1\) (from the \(z-1\) factor) and \(z=\pm 2i\) (from the \(z^{2}+4\) factor).

Now, let's compute the constants \(A\), \(B\), and \(C\) in the partial fractions decomposition: $$ \frac{4z^{2}-2z+8}{(z-1)(z^{2}+4)}=\frac{A}{z-1}+\frac{Bz+C}{z^{2}+4}. $$ Multiply both sides by the common denominator \((z-1)(z^{2}+4)\): $$ 4z^{2}-2z+8=A(z^{2}+4)+(Bz+C)(z-1). $$ We will solve the coefficients by substituting the pole values: For the pole \(z=1\), the equation becomes: $$ 4(1)^2-2(1)+8=A(1^2+4). $$ $$ A=\frac{10}{5}=2. $$ For the pole \(z=\pm2i\), we have: $$ 4(-4)+8 = (B(2i)+C)((\pm2i)-1). $$ Simplifying the equation above: $$ -8 = (\pm2Bi - B + Ci - C)(\pm2i-1), $$ which yields two equations for the real and imaginary parts: $$ 0 = -B-C, \\ -8 = 2Bi-C(2i), $$ Solving the system of equations, we find that \(B=-4\) and \(C=4\). Therefore, our partial fraction decomposition is: $$ f(z) = \frac{2}{z-1} + \frac{-4z+4}{z^{2}+4}. $$ Now to compute the residues: Residue at \(z=1\): Residue \(=\lim_{z\to 1} (z-1)\frac{2}{z-1} = 2\). Residues at \(z=\pm 2i\): There is no simple pole at \(z=\pm 2i\), so no residues are computed. To summarize: 1. The partial fraction decomposition is \(f(z) = \frac{2}{z-1} + \frac{-4z+4}{z^{2}+4}\) 2. Poles are at \(z=1\) and \(z=\pm 2i\) 3. Residues at the poles are \(2\) at \(z=1\) and no residues at \(z=\pm 2i\).