A question on a contour integral Let $C$ denote the circle, say, for any fixed $R>0$,
$z=Re^{i\theta}, \ (-\pi\leq\theta\leq\pi)$
on the complex plane. To evaluate the complex integral $\int_{C}\frac{1}{z} dz$ we cannot use an antiderivative of the function $\frac{1}{z}$. Because, if we denote by $F(z)$ a branch of $\log z$, then the derivative of $F(z)$ is $\frac{1}{z}$, but $F(z)$ is not differentiable along its branch cut.
 However we can make use of an antiderivative in the following way:
 Let $\epsilon>0$ and let $C_{\epsilon}$ denote the arc
$z=Re^{i\theta}, \ (-\pi+\epsilon\leq\theta\leq\pi-\epsilon)$,
and let $z_1^{\epsilon}$ and $z_2^{\epsilon}$ denote, respectively, the initial and final points of the arc $C_{\epsilon}$. Then the principal branch
$Log z=\ln |z|+i\Theta, \ (|z|>0, \ -\pi<\Theta<\pi)$
of the logarithmic function can be used as an antiderivative of the function  $\frac{1}{z}$ to evaluate the integral $\int_{C_{\epsilon}}\frac{1}{z} dz$. Indeed, since $z_1^{\epsilon}=Re^{i(-\pi+\epsilon)}$ and  $z_2^{\epsilon}=Re^{i(\pi-\epsilon)}$, we get
$\int_{C_{\epsilon}}\frac{1}{z} dz=\int_{z_1^{\epsilon}}^{z_2^{\epsilon}}\frac{1}{z} dz=Log|_{z_1^{\epsilon}}^{z_2^{\epsilon}}=Log(z_2^{\epsilon})-Log(z_1^{\epsilon})=2\pi i-2\epsilon$.
Now, since $\epsilon$ is arbitrary, we obtain
(1)  $\int_{C}\frac{1}{z} dz\stackrel{(*)}=\lim_{\epsilon\longrightarrow 0}\int_{C_{\epsilon}}\frac{1}{z} dz=\lim_{\epsilon\longrightarrow 0}(2\pi i-2\epsilon)=2\pi i$,
which is correct value of the integral $\int_{C}\frac{1}{z} dz$. My question/doubt here is the following: In the statement (1), is the first equality $\stackrel{(*)}=$ correct/legitimate ? If so, is there any general result (theorem, proposition, etc.) that guarantees this kind of equality, and is there any reference in this direction ?
ADDED: The reason for my question is the following:  Of course, there are several ways to find the value of the above integral. However, here I would like to use just an antiderivative to find the value of the given integral. So, if we do not consider the branch cut of $Log z$ (of course, we must consider!), then by taking $z_1=Re^{i(-\pi)}$ and $z_2=Re^{i(\pi)}$, formally we get  
$\int_{C}\frac{1}{z} dz=\int_{z_1}^{z_2}\frac{1}{z} dz=Log|_{z_1}^{z_2}=Log(z_2)-Log(z_1)=2\pi i$,
which is the same value of the integral; however it is not legitimate to use the $Logz$ as an antiderivative, as $Logz$ does not have a derivative at the point $z=-R$. Now, I am trying to understand that why in this way we got the correct value of the integral. Is it just an coincidence, or it does rely on a general result, or it is a consequence of a general result? And, I wonder if there is an example of an integral such that when we computing the value of the integral using an antiderivative in an incorrect way as above, we get a value which is not a correct/real value of the integral !! 
 A: All you need is the fact that $\frac 1 z$ is bounded on $C\setminus C_{\epsilon}$ (with a bound indepdent of $\epsilon$). The length of $C\setminus C_{\epsilon}$ tends to $0$ so the integral over this path tends to $0$. 
A: The function $\frac{1}{z}$ is of the form $\frac{f(z)}{z-z_o}$ where $f(z)=1~$ and $z_0 = 0$.
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Now, the circle $C$ is a closed contour and $\frac{1}{z}$ is analytic everywhere inside and on the circle except the point $z_0 = 0$. Under these circumstances Cauchy's Integral Formula;
$$f(z_0) = \frac{1}{2\pi i} \int_C \frac{f(z)}{z-z_0}~dz$$ is valid and for this particular case we have:$$\int_C \frac{1}{z}~dz = 2\pi i\cdot f(0) = 2\pi i$$
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The proof for Cauchy's integral formula involves splitting the integral into a sum of two integrals over a circle contour and it's possible to take the limit of one of these integrals and let the radius tend to zero. Unlike your limit, this limit equals zero, but that's the key in proving Cauchy's integral formula.
I don't know if Cauchy's formula is a reference in the direction you had in mind but your method can be used to show that $\int_C \frac{1}{z-z_0}~dz = 2\pi i$ for any $z_0$ inside $C$, the general theorem is:
$$\oint_C(z-z_o)^n~dz = \begin{cases}0 ~~~~~~~~~~if~~ n \neq -1\\2\pi i ~~~~~~if~~ n = -1\end{cases}$$
