Discussion
Let \(C\) be the ellipse \(\{z \in \mathbb{C} : |z - 2| + |z + 2| = 8\}\) traversed counter-clockwise. The value of the contour integral
\[ \oint_{C} \frac{z^{2}\,dz}{z^{2} - 2z + 2} \]
is equal to
Answer: C
The integral equals \(4\pi i\).
The poles are the roots of \(z^{2} - 2z + 2 = 0\), namely \(z = 1 \pm i\). For \(z = 1 + i\), \(|z - 2| + |z + 2| = \sqrt{2} + \sqrt{10} \approx 4.58 \lt 8\), so both poles (they are symmetric) lie inside the ellipse.
Write the integrand as \(1 + \frac{2z - 2}{z^{2} - 2z + 2}\). The constant integrates to zero, and the second term is \(q'/q\) for \(q = z^{2} - 2z + 2\), whose integral is \(2\pi i\) times the number of zeros of \(q\) inside \(C\):
\[ \oint_{C} = 0 + 2\pi i \times 2 = 4\pi i. \]
\(2\pi i\) counts only one pole, and 0 would hold only if both poles were outside the ellipse.
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