Discussion

The value of the integral

\[ \oint(\frac{6z}{2z^{4}-3z^{3}+7z^{2}-3z+5})dz \]

evaluated over a counter-clockwise circular contour in the complex plane enclosing only the pole \(z = i\), where \(i\) is the imaginary unit, is

Answer: A

The integral equals \((-1 + i)\pi\).

The denominator factorises, as substituting \(z = i\) suggests (it gives \(2 + 3i - 7 - 3i + 5 = 0\)):

\[ \begin{aligned} &2z^{4} - 3z^{3} + 7z^{2} - 3z + 5 \\ &= (z^{2} + 1)(2z^{2} - 3z + 5) \end{aligned} \]

So \(z = i\) is a simple pole, and its residue is

\[ \begin{aligned} \text{Res} &= \frac{6i}{(2i)(2i^{2} - 3i + 5)} = \frac{3}{3 - 3i} \\ &= \frac{1}{1 - i} = \frac{1 + i}{2} \end{aligned} \]

By the residue theorem the integral is \(2\pi i \times \frac{1 + i}{2} = \pi(i - 1)\).

\((1 + i)\pi\) is the residue times \(2\pi\) without the factor \(i\), and the other two options do not come from this residue.

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