Average value of a complex valued function on a circle. The following is an exercise from Complex Analysis by Stephen Fisher.
Fix a complex number $a$ and a positive real number $R$. Suppose $u$ is a function defined on the circle of radius $R$ centered at $a$. Let $C$ denote this circle. 
Show that the average value of $u$ on $C$ is given by $\frac{1}{2\pi}\int_{0}^{2\pi} u(a + Re^{it})dt$.
Any Hints please.
 A: We assume Cauchy's integral theorem.
$$(1)\hspace{5mm}\frac{1}{2\pi}\int_{0}^{2\pi} u(a+ Re^{i t})dt = \frac{1}{2\pi i}\int_{0}^{2\pi}\frac{u(a+Re^{it})iRe^{it}}{Re^{it}}dt $$
The right side of (1), letting R be the radius of circle $C,$  $u(z)$ the equation of the circle $|z-a|= R $ or $z = a + Re^{it},$ so that $z-a = Re^{it}$ and $dz = iRe^{it},$ is precisely 
$$(2)\hspace{5mm}\frac{1}{2\pi i}\oint_C\frac{u(z)}{z-a}dz$$ 
By Cauchy's integral formula (2) is equal to $u(a).$ $\square  $
This exercise is odd because the starting integral is generally taken as the definition of the average value, as noted in a comment above.  We would normally derive that form for the average from Cauchy's integral formula.  
A: As implied in the comments, the most important part in this exercise is the interpretation of "average value". As always, in order to define the average value of a function on $C$ we need a measure on $C$. One should note that different measures evidently yield different average values. However, as $C$ is a smooth curve, it makes sense to take the standard length measure.
Parametrize $C$ by$$\gamma:[0,2\pi R]\to\mathbb{C},\quad s\mapsto a+Re^{is/R}.$$This is an arc-length parametrization, or in other words, pulling back the length measure on $C$ with respect to $\gamma$ yields the standard Lebesgue measure on $[0,2\pi R]$. Hence, the average value of $u$ on $C$ is equal to the average value of $u\circ\gamma$ on $[0,2\pi R]$. The latter is just$$\frac{1}{2\pi R}\int_0^{2\pi R}u\left(a+Re^{is/R}\right)ds.$$The desired expression follows now by substituting $t=s/R$. 
