Morera's Theorem and independence of path Determine whether 
$$
G(z)=\int_{1}^{z}\frac{1}{t} dt
$$
is independent of the path joining $1$ and $z$ , and discuss the relationship of your answer to Morera's Theorem .....  I believe that as long as the paths goes through a region that excludes the origin ,the integral is independent of the path.... but I am not sure if this is correct or what type of argument is required here , in particular the relation of the answer to Morera's theorem is not clear to me .
 A: Have you thought about traversing the unit circle for $z=1$?
$$
\int_C\frac 1zdz=\int_0^{2\pi}\frac{1}{e^{it}}(ie^{it})dt=\int_0^{2\pi}idt=2\pi i
$$
A: some thoughts may be of assistance:
suppose we have a circuit $\partial \Gamma$ which encloses a simply-connected open region $\Gamma$ which excludes the origin. 
on $\Gamma \cup  \partial \Gamma$ we  have:
$$
\frac{dz}{z} = \frac{(x-iy)(dx+idy)}{x^2+y^2} =\frac{(xdx+ydy)+i(xdy-ydx)}{x^2+y^2}
$$
dealing with the real and imaginary parts by turn, Green's theorem shows that the integral around $\partial \Gamma$ is zero.
set $L=\frac{x-iy}{r^2}$,  and $M=\frac{y+ix}{r^2}$
then
$$
\frac{\partial L}{\partial y} = -\frac{i}{r^2}-\frac{2(x-iy)y}{r^4}
$$
and
$$
\frac{\partial M}{\partial x} = \frac{i}{r^2}-\frac{2(y+ix)x}{r^4}
$$
thus
$$
\int_{\partial \Gamma} \frac{dz}{z} = \int_{\partial \Gamma}Ldx+Mdy=\int_{\Gamma} (\frac{\partial M}{\partial x}-\frac{\partial L}{\partial y}) dx dy \\
$$
but, as is easily checked,
$$
\frac{\partial M}{\partial x}-\frac{\partial L}{\partial y} = 0
$$
