The sum of residues of fz-2zz-12z-2 at its singular point is

F(z)=2z(z 1)^2(z 2) Poles are z=2 (simple pole) and z=1 (double pole) R1=Resf(z)(z nbsp;=2)=limz→2(z 2)f(z)=limz→22z(z 1)2=4R2=Resf nbsp;(z)(z=1)=1(2 1!)ddz ...

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Standard XII

Mathematics

Definition of Circle

The sum of re...

Question

The sum of residues of $f (z) = \frac{2 z}{(z - 1)^{2} (z - 2)}$ at its singular point is

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Solution

The correct option is C $0$
$f (z) = \frac{2 z}{(z - 1)^{2} (z - 2)}$
Poles are $z = 2$ (simple pole) and $z = 1$ (double pole)
$R_{1} = R e s f (z) (z = 2) = lim z \to 2 (z - 2) f (z) = lim z \to 2 (\frac{2 z}{(z - 1)^{2}}] = 4$ $R_{2} = R e s f (z) (z = 1) = \frac{1}{(2 - 1!)} {(\frac{d}{d z} {(z - 1)}^{2} f (z)]}_{z = 1}$ $= {[\frac{d}{d z} (\frac{2 z}{z - 2})]}_{z = 1} = - 4$
Hence sum of residues $= R_{1} + R_{2} = 0$

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The sum of residues of f(z)=2z(z−1)2(z−2) at its singular point is

The correct option is C 0f(z)=2z(z−1)2(z−2) Poles are z=2 (simple pole) and z=1 (double pole) R1=Resf(z)(z =2)=limz→2(z−2)f(z)=limz→2(2z(z−1)2]=4R2=Resf (z)(z=1)=1(2−1!)(ddz(z−1)2f(z)]z=1=[ddz(2zz−2)]z=1=−4 Hence sum of residues=R1+R2=0

The sum of residues of $f (z) = \frac{2 z}{(z - 1)^{2} (z - 2)}$ at its singular point is