Discussion
Let \(\int_{C} \left[\frac{1}{(z-2)^{4}} - \frac{(a-2)^{2}}{z} + 4\right] dz = 4\pi\), where the close curve \(C\) is the triangle having vertices at \(i\), \(\left(\frac{-1-i}{\sqrt{2}}\right)\) and \(\left(\frac{1-i}{\sqrt{2}}\right)\), the integral being taken in anti-clockwise direction. Then one value of \(a\) is
Answer: C
One value is \(a = 3 + i\).
The three vertices lie on the unit circle at angles \(90^{\circ}\), \(225^{\circ}\) and \(315^{\circ}\), so the triangle encloses the origin but not \(z = 2\). The term \(1/(z - 2)^{4}\) is analytic inside \(C\) and the constant 4 integrates to zero, so only the pole at 0 contributes:
\[ \begin{aligned} -(a - 2)^{2}\cdot 2\pi i &= 4\pi \\ (a - 2)^{2} &= \frac{4\pi}{-2\pi i} = 2i \\ a - 2 &= \pm(1 + i) \end{aligned} \]
since \((1 + i)^{2} = 2i\). So \(a = 3 + i\) or \(a = 1 - i\), and only \(3 + i\) is offered. \(1 + i\), \(2 + i\) and \(4 + i\) give \((a-2)^{2} = -2i\), \(-1\) and \(3 + 4i\), none equal to \(2i\).
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