The residues of a function fz-1z-4z-13

F(z)=1(z 4)(z+1)^3 Residue at z=4: nbsp; lim_z → 4[(z 4)f(z)] =1(5)^3=1125 Residue at z= 1: lim_z → 112!d^2dz^2 [ (z+1)^3×1(z 4)(z+1)^3 ] =12!×2( 5)^3= 1125 ...

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Engineering Mathematics

Type of Singularities and Residues

The residues ...

Question

The residues of a function $f (z) = \frac{1}{(z - 4) (z + 1)^{3}}$

\frac{- 1}{27}

and

\frac{- 1}{125}

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\frac{1}{125}

and

\frac{- 1}{125}

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\frac{- 1}{27}

and

\frac{1}{5}

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\frac{1}{125}

and

\frac{- 1}{5}

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Solution

The correct option is B $\frac{1}{125}$ and $\frac{- 1}{125}$
$f (z) = \frac{1}{(z - 4) (z + 1)^{3}}$
Residue at z=4:
${lim}_{z \to 4} [(z - 4) f (z)]$
$= \frac{1}{(5)^{3}} = \frac{1}{125}$
Residue at z=-1:
$l i m_{z \to - 1} \frac{1}{2!} \frac{d^{2}}{d z^{2}} [(z + 1)^{3} \times \frac{1}{(z - 4) (z + 1)^{3}}]$
$= \frac{1}{2!} \times \frac{2}{(- 5)^{3}} = \frac{- 1}{125}$
So residues are $\frac{1}{125}$ and $\frac{1}{125}$

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The residues of a function f(z)=1(z−4)(z+1)3

The correct option is B 1125 and −1125f(z)=1(z−4)(z+1)3 Residue at z=4: limz→4[(z−4)f(z)] =1(5)3=1125 Residue at z=-1: limz→−112!d2dz2[(z+1)3×1(z−4)(z+1)3] =12!×2(−5)3=−1125 So residues are 1125 and 1125

The residues of a function $f (z) = \frac{1}{(z - 4) (z + 1)^{3}}$