Discussion
Let \(f\) be analytic on \(\mathbf{C} - \{0\}\) and let \(\nu_{1}, \nu_{2} : [0, 1] \to \mathbf{C} - \{0\}\) be defined by \(\nu_{1}(t) = e^{2\pi i t}\) and \(\nu_{2}(t) = 10\, e^{2\pi i t}\). Suppose \(I_{1} = \int_{\nu_{1}} f(z)\, dz\) and \(I_{2} = \int_{\nu_{2}} f(z)\, dz\).
Then :
Answer: A
The answer is \(\mathrm{I}_{1} = \mathrm{I}_{2}\).
Both curves go once anticlockwise around the origin: \(\nu_{1}\) is the unit circle and \(\nu_{2}\) the circle of radius 10. They are homotopic in \(\mathbf{C} - \{0\}\), through the circles of radius \(r\) between 1 and 10, and \(f\) is analytic on the annulus between them. By Cauchy's theorem (the integral over the boundary of the annulus is zero), the two integrals are equal.
The radius does not scale the integral. For \(f(z) = 1/z\), both integrals equal \(2\pi i\), which rules out \(\mathrm{I}_{1} = 10\,\mathrm{I}_{2}\) and \(\mathrm{I}_{1} = \mathrm{I}_{2}/10\), and the relation is a definite one, not 'no relation'.
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