The value of -r3z-5z-1z-2dz along a closed path - is equal ot 4- i- where z-x-iy and i-1-The correct - is

Method I: Let f(z)=3z 5(z 1)(z 2) so poles are z=1 amp; 2 R1=Res nbsp;f(z)(z=1)=limz→1((z 1)f(z))=2R1=Res nbsp;f(z)(z=2)=limz→2((z 2)f(z))=1Let us assume ...

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Engineering Mathematics

Cauchy Integral Formula and Problems

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Question

The value of $\oint_{r} \frac{3 z - 5}{(z - 1) (z - 2)} d z$ along a closed path $Γ$ is equal ot $4 π i$ , where $z = x + i y$ and $i = \sqrt{- 1}$ .
The correct $Γ$ is

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Solution

The correct option is B
Method I:
Let $f (z) = \frac{3 z - 5}{(z - 1) (z - 2)}$ so poles are $z = 1$ & $2$
$R_{1} = R e s f (z) (z = 1) = l i m z \to 1 ((z - 1) f (z)) = 2$ $R_{1} = R e s f (z) (z = 2) = l i m z \to 2 ((z - 2) f (z)) = 1$ Let us assume that $z = 1$ lies inside given contour and $z =$ lies outisde it.
Then by C - R - T,
$\oint f (z) d z = 2 π i$ (sum of residues at poles inside)
$= 2 π i (2) = 4 π i$
Hence verified,
So our assumption that $z = 1$ lies inside $Γ$ and $z = 2$ lies outside $Γ$ so correct path of $Γ$ is defined by (b)

Method II:
Using Cauchy's formula,
$\oint \frac{3 z - 5}{(z - 1) (z - 2)} d z = \oint \frac{\frac{3 z - 5}{z - 2}}{z - 1} d z$
$= j 2 π {[\frac{3 z - 5}{z - 2}]}_{a t z = 1}$
$= j 2 π [\frac{3 - 5}{1 - 2}]$
$= j 2 π [\frac{- 2}{- 1}] = j 4 π$
Hence, $z = 1$ lies inside the closed path $Γ$ . and $z = 2$ lies outside the closed path $Γ$ .

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The value of ∮r3z−5(z−1)(z−2)dz along a closed path Γ is equal ot 4πi, where z=x+iy and i=√−1. The correct Γ is

The value of $\oint_{r} \frac{3 z - 5}{(z - 1) (z - 2)} d z$ along a closed path $Γ$ is equal ot $4 π i$ , where $z = x + i y$ and $i = \sqrt{- 1}$ .
The correct $Γ$ is