Multiple choice

The residues of a complex function X(z) = $\frac{1 - 12z}{z(z - 1)(z - 2)}$at its poles are

  1. $\frac{1}{2}$, $\frac{1}{2}$ and 1
  2. $\frac{1}{2}$, $\frac{1}{2}$and – 1
  3. $\frac{1}{2}$, 1 and –3/2
  4. $\frac{1}{2}$, – 1 and $\frac{2}{3}$
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C Correct answer
Explanation

 X(z)= (1-2z)/{z(z-1)(z-2)}      Poles are located at z=0, z=1 and z=2          At z=0            = z(1-2z)/{z(z-1)(z-2)} = 1/2          At z=1            =(z-1)(1-2z)/{z(z-1)(z-2)}= 1            At z=2           =(z-2)(1-2z)/{z(z-1)(z-2)} = -3/2      

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