Show that there is no branch of $\arg z$ in $\{0 \lt |z| \lt 1 \}$ using the fact that $\int_{\gamma} \frac{dz}{z}=2\pi i$ , where $\gamma(t)=Re^{it}, o \le t\le 2\pi$, $R\gt 0$.
Suppose there is a branch of $\arg z$ in the mentioned domain say ($\alpha(z)$). Then there would have been a branch $\beta$ of logarithm given by $\beta(z)=\ln|z|+i \alpha(z)$. Then $e^{\beta(z)}=z$.
Then I should be able find a Holomorphic Function $F(z)$ such that $F'(z)=f(z)=\frac{1}{z}$ and that would give me the value of integration to be $0$ as $\gamma(0)=\gamma(1)$.
But what should that $F(z)$ be??
Thanks for the help!!
We don't need the power of complex analysis in order to deal with this problem. Note that the branches of ${\rm arg}$ are real-valued in the first place. We can argue as follows instead: If $$\alpha:\quad\Omega\to{\mathbb R},\qquad (x,y)\mapsto\alpha(x,y)$$ is a branch of ${\rm arg}$ in some domain $\Omega\subset\dot{\Bbb R}^2$ then $$\nabla\alpha(x,y)=\left({-y\over x^2+y^2}, \>{x\over x^2+y^2}\right)\qquad\bigl((x,y)\in\Omega\bigr)\ ,$$ and we should have $$\int_\gamma \nabla\alpha \cdot d{\rm z}=0\tag{1}$$ for all closed curves in $\Omega$. Now assume that $\Omega$ is the punctured disc of radius $1$, and let $$\gamma:\quad t\mapsto{1\over2}(\cos t, \>\sin t)\qquad(0\leq t\leq2\pi)$$ be the circle of radius ${1\over2}$. Computing the integral $(1)$ for this circle gives $2\pi$, of course. This rules out the possibility of a univalent branch of $\arg$ for this $\Omega$.