Cauchy Integral formula with central angle different from $2\pi$

Question

We know that if $f$ is holomorphic in an open set $A$ and continous in its closure, and $C$ is a closed curve inside $A$ then :

$f(z_0)2i\pi = \int_{C} \frac{f(z)}{z- z_0} dz $

Now suppose that $C$ its not closed, that is, the central angle of C is not $2\pi$. Lets call the central angle of the arc $\theta$

Can we say that $f(z_0)\theta i\pi = \int_{C} \frac{f(z)}{z- z_0} dz $ ?

Edit. I will add some context to que question :

I am trying to prove that if $z_0$ is a simple pole of $f$ And if $C$ is an arc of radius $r$ and angle $\theta$ around $z_0$, then :

$Res(f, z_0) = \lim_{r\to 0^+}\frac{1}{i\theta}\int_{C} f(z) dz $

Answer 1

No, we can't. There is no reason for $\int_C \frac{f(z)}{z-z_0}\; dz$ to have anything to do with $f(z_0)$.

EDIT: Hint for what you're trying to prove: note that if $z_0$ is a simple pole, $$ f(z) = \frac{\text{Res}(f,z_0)}{z-z_0} + g(z) $$ where $g(z)$ is holomorphic in a neighbourhood of $z_0$.

Answer 2

$\def\res{\mathop{\text{Res}}\limits}$ In case of simple pole, the limit can be proved, but the quantities $i\theta f(z_0)$ and $\int_C f(z)dz$ do not coincide in general. If $f$ has a simple pole, we may write $$f(z) = \frac{B}{z-z_0}+g(z)$$ where $g$ is holomorphic at $z_0$. In particular $g$ is bounded near $z_0$, so that the integral of $g$ over $C$ vanishes as $r\to 0+$. Remaining part is $$\int_C\frac{B}{z-z_0}dz = \int_{\theta_0}^{\theta_0+\theta}\frac{B}{re^{it}}rie^{it}dt=i\theta B,$$ so we have $$\lim_{r\to 0+}\int_Cf(z)dz = i\theta\res_{z=z_0}f(z).$$ The vanishing(as $r\to 0+$) term $\int_C g(z)dz$ is the difference of two quantities.