Question

The residuses of a complex function $X\left(z\right)=\frac{1-2z}{z(z-1)(z-2)}$ at its poles are

• A. $\frac{1}{2},-\frac{1}{2}$ and $1$
• B. $\frac{1}{2},-\frac{1}{2}$ and $-1$
• C. $\frac{1}{2},1$ and $-\frac{3}{2}$
• D. $\frac{1}{2},-1$ and $\frac{3}{2}$

Answer 1

The correct option is C $\frac{1}{2},1$ and $-\frac{3}{2}$

$X\left(z\right)=\frac{1-2z}{z(z-1)(z-2)}$
$z=0,1,2$ ase simple poles.
$\left[Resx\left(z\right);0\right]$
$=limz\to 0\frac{Z(1-2z)}{z(z-1)(z-2)}$
$=\frac{1}{(-1)(-2)}=\frac{1}{2}$
$\left[Resx\left(z\right);1\right]$
$=limz\to 1\frac{(z-1)(1-2z)}{z(z-1)(z-2)}$
$=\frac{1}{1(-1)}=1$
$\left[Resx\left(z\right);2\right]$
$=limz\to 2\frac{(z-2)(1-2z)}{z(z-1)(z-2)}$
$=\frac{1-4}{2(2-1)}=-\frac{3}{2}$