Does the closed curve achieves its maximum/minimum? This question is aimed at the complex plane.
Let $\gamma$ be a closed curve and suppose that $a$ is not a point on the curve $\gamma$ and I wanted to consider the integral $\int_\gamma \frac{dz}{z-a}.$
So am I right in saying that $\frac{1}{z-a}$ is holomorphic on $\gamma$? Also can I bound $\frac{1}{z-a}$ by $M$ on $\gamma$ so that $\int_\gamma \frac{dz}{z-a}\leq M|\gamma|$ where the modulus of gamma is the length of curve? How do I know if our function is bounded? I can see that $\gamma$ is bounded because it is compact but I have no idea how that may help.
Also am I right in claiming that the length of any closed curve is finite, does it follow straight from the definition or does this require some form of proof?
 A: I don't know how much topology you're familiar with, but a closed curve will contain any of its limit points; in geometric terms, this means that the distance to any point not on the curve is well-defined and strictly positive (this would not be the case for 'open curves'). This should fairly easily lead to the upper bound $M$, using the fact that the modulus of the integral is smaller or equal to the integral of the modulus of the integrand.
Note that $\gamma$ is not compact per se. Compact implies closed (in this topology); closed does not imply compact.
A: $\gamma$ is compact as it is the continuous image of $[0,1]$. Since $a$ is not on $\gamma$, the distance from $a$ to $\gamma$ is positive. Indeed, the function $|z-a|$ is positive on $\gamma$, and by compactness achieves its minimum $m$ at some point $w$ on $\gamma$, that is $m=|w-a|\le|z-a|$ for all $z$ on $\gamma$. Then if we let $M=\frac1m$ we have $\frac1{|z-a|}\le M$ for all $z$ on $\gamma$. 
$|\gamma|$ need not be finite. If $|\gamma|$ is finite then indeed $\int_\gamma \frac{dz}{z-a}\le M|\gamma|$. But it could be that $|\gamma|=\infty$. Nevertheless $\int_\gamma \frac{dz}{z-a}$ is finite, e.g. if $\gamma$ is a simple (no self-intersections) closed curve (so then "inside" would be well-defined), and if $a$ is not "inside" the curve $\gamma$ (i.e. if $a$ does not belong to the bounded component of the complement of $\gamma$) then $\int_\gamma \frac{dz}{z-a}=0$. If $\gamma$ is a simple closed curve and $a$ is inside $\gamma$ then $\int_\gamma \frac{dz}{z-a}=\pm2\pi i$ (depending on the direction/orientation of $\gamma$). If $\gamma$ is closed, but not necessarily simple, then $\int_\gamma \frac{dz}{z-a}=k2\pi i$ for some integer $k$ depending on how many times and in what direction does $\gamma$ "circle around" $a$. (So, the $M|\gamma|$ is irrelevant anyway, unless you are interested in $\int_\gamma \frac{dz}{|z-a|}$ instead of $\int_\gamma \frac{dz}{z-a}$. If $|\gamma|=\infty$ then $\int_\gamma \frac{dz}{|z-a|}=\infty$ too, even though 
$\int_\gamma \frac{dz}{z-a}=k2\pi i$ is finite.) 
For an example of a differentiable curve of infinite length (using $(x,y)$ instead of $z$), let $f(x)=x^2\sin\frac1{x^5}$, $f(0)=0$, and $-1\le x\le1$. The idea is that the $x^2$ term makes this function differentiable at $0$ (and the derivative at $0$ is $0$), on the other hand the $\frac1{x^5}$ term makes the curve oscillate so frequently near $0$ that it "accumulates" infinite length 
(and it would have infinite length for any interval $-\varepsilon\le x \le\varepsilon$). (I didn't verify the details, most likely $\frac1{x^5}$ is an overkill, perhaps $\frac1{x^2}$ would have sufficed, on the other hand just to stay on the safe side one may take $\frac1{x^{100}}$ should it be necessary.)  
