TITLE: Problem evaluating a contour integral using parametrization
QUESTION [7 upvotes]: I tried to solve the following contour integral:
$$
\oint_\gamma  {\frac{{dz}}{{z - c}}}
$$
Where $\gamma$ is a disk centered at the origin. In order to do so, I used the following parametrization: 
$$
\begin{array}{l}
 z &= Re^{i\varphi }, \qquad 0 < \left| R \right| \ne \left| c \right|  \\ 
 dz &= iRe^{i\varphi } d\varphi 
 \end{array}
$$
Replacing in the contour integral:
$$
\begin{array}{l}
 \oint_\gamma  {\frac{{dz}}{{z - c}}}  &= \int\limits_0^{2\pi } {\frac{{iRe^{i\varphi } }}{{Re^{i\varphi }  - c}}} d\varphi  \\ 
  &= \left. {\ln \left( {Re^{i\varphi }  - c} \right)} \right|_0^{2\pi }  \\ 
  &= \ln \left( {Re^{i2\pi }  - c} \right) - \ln \left( {Re^{i0}  - c} \right) \\ 
  &= \ln \left( {R - c} \right) - \ln \left( {R - c} \right) \\ 
  &= 0
 \end{array}
$$
However, by the residue theorem the contour integral must be equal to $2\pi i$ if $\left| R \right| > \left| c \right|$, whereas in the answer obtained by parametriztion the value is always $0$.
My question is: What am I missing here? Where is my mistake? Thank you in advance.

REPLY [3 votes]: Because there is no branch of logarithm in $\mathbb{C}\setminus \{0\}$, so your function has no antiderivative. That's a big difference between real and complex analysis-you can't take the elementary real functions and just use them in the complex plane. Logarithm is a function which requires a branch.