# Centre of gravity of image of circle in the complex plane (Whittaker & Watson)

Whittaker & Watson Ch. 6. Problem 7. is a result apparently due to E. Amigues, that if $C$ is a circle in the complex plane centred at $a$, $z\in C,$ and the function $f$ is analytic on and inside $C$ except at some number of poles, then "if to each element of $\gamma$ be attributed a mass proportional to the corresponding element of $C,$ the centre of gravity of $\gamma$ is the point $r,$ where $r$ is the sum of the residues of $f(z)/(z-a)$ at its poles in the interior of $C.$"

I can sort of intuitively see where this result is coming from, since if $f$ is regular then $r=f(a)$ follows immediately from Cauchy's formula, and I suppose the idea is that $f$ just twists the circle around a bit in such a way that the centre of the circle is mapped into the centre of gravity of the curve $\gamma,$ but I'm not quite sure how to explicitly go about showing that the centre of gravity should be the sum of the residues of $f(z)/(z-a).$

Just looking for a helping hand. Thanks.

How I interpreted the question was as a small element of mass $dm$ of the curve $\gamma$ is essentially equivalent to $d\theta,$ where $\theta=\arg z,$ up to some constant of proportionality which we might as well let be unity and it won't matter anyway.
Then the total mass of the curve $\gamma$ can be said to be $2\pi,$ and then we have the centre of gravity given by
$$\frac{1}{2\pi}\oint_\gamma f(z)dm=\frac{1}{2\pi i}\oint_C\frac{f(z)dz}{z-a}=r$$