Using partial fractions dfrac 5Z-2 Z(Z-1) =dfrac 2 Z +dfrac 3 Z-1 , the value of displaystyleoint_ |Z|=2 dfrac 5Z-2 Z(Z-1) ,dZ by the Residue Theorem is:

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SSC Pure Mathematics QUESTION #4100

Question 1

Using partial fractions \(\dfrac{5Z-2}{Z(Z-1)}=\dfrac{2}{Z}+\dfrac{3}{Z-1}\), the value of \(\displaystyle\oint_{|Z|=2}\dfrac{5Z-2}{Z(Z-1)}\,dZ\) by the Residue Theorem is:

\(2\pi i\)
\(4\pi i\)
\(10\pi i\)✔️
\(6\pi i\)

Correct Answer Explanation

Both poles \(Z=0\) (residue \(2\)) and \(Z=1\) (residue \(3\)) lie inside \(|Z|=2\). By Cauchy's Residue Theorem: \(\oint = 2\pi i(\text{Res}_{Z=0}+\text{Res}_{Z=1})=2\pi i(2+3)=10\pi i\).

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