Showing a complex function defined as an integral is holomorphic 
Let $\gamma:[0,1] \to \mathbb{C}$ be any $C^1$ curve. Define $$f(z)=\oint_\gamma \frac{1}{\zeta-z}d\zeta.$$
  Prove that $f$ is holomorphic on $U=\mathbb{C}-\hat{\gamma}$ where $\hat{\gamma}=\{ \gamma(t):0\leq t\leq 1 \} .$ In case $\gamma(t)=t,$ show that there is no way to extend $f$ to a continuous function on all of $\mathbb{C}$. 

I know in my head we're really discussing the complex logarithm here, and in the second part in specific we are discussing, say, $Log(\frac{z}{z-1})$ which has $[0,1]$ as its branch cut (not sure if I am using the terminology correctly here). My initial thoughts on the first part are to somehow use Morera's Theorem or simply to use the definition of differentiation, but I can't sort out how the details would work. 
 A: For the second part of the question, if $\gamma(t)=t$ and $z=x+iy$ is not on the positive real axis, then 
$f(z)=\oint_\gamma \frac{1}{\zeta-z}d\zeta=\int_{0}^{1}\frac{1}{(t-x)-iy}dt=\int_{0}^{1}\frac{(t-x)+iy}{(t-x)^2+y^2}dt$
$=\int_{0}^{1}\frac{(t-x)}{(t-x)^2+y^2}dt + iy\int_{0}^{1}\frac{1}{(t-x)^2+y^2}dt$
$=(1/2)\int\frac{ds}{s}ds + (i/y)\int_{0}^{1}\frac{1}{[(t-x)/y]^2+1}dt$ for $s=(t-x)^2+y^2$.
$=(1/2)ln|s|+iarctan((t-x)/y)$ evaluated over the interval $[0,1]$.
Back substituting the expression for $s$, one sees that the final number is 
$f(z)=(1/2)ln|1-2x+r^2|-ln|r|+i[arctan((1-x)/y)+arctan(x/y)]$,
where $r^2$ as usual is $x^2 + y^2$. 
Now suppose $f$ extends. Taking $z=cos\tau+isin\tau$, which is a parameterization of the unit circle, we have for $\tau \neq n\pi$,
$f(z(\tau))=(1/2)ln|2-2cos\tau|+i[arctan((1-cos\tau)/sin\tau)+arctan(cot\tau)]$,
Now since we believe that $f$ is continuous over all of $\mathbb{C}$, the functions corresponding to its real and imaginary parts also extend continuously over $\mathbb{C}$. In particular, the function
$g(\tau):=Im(f(z(\tau)))= arctan((1-cos\tau)/sin\tau)+arctan(cot\tau)$,
where this equality holds for $\tau \neq n\pi$ extends to a continuous function of $\tau$. But this cannot be since the limit of $g$ as $\tau$ approaches 0 from the left is $arctan(-\infty)+arctan(-\infty)=-\pi$. On the other hand, the limit as $\tau$ approaches 0 from the right is $arctan(\infty)+arctan(\infty)=\pi$, so there's no hope of finding a value of $g$ at $\tau=0$ (hence, of $f$ at $z=1$) that makes this function continuous. 
Note that the imaginary parts differ on opposite sides of $\gamma$ by $2\pi$. This is as expected.
