Contour integral $\int_{C}Log(z+3)$ I am trying to compute $\int_{C}Log(z+3)$, where C is a circle centered at the origin, with radius of 2, oriented once counterclockwise. 
I kind of get the idea of how to compute $\int Log(z)$ on the unit circle. I know $z=e^{i \theta}$ and $dz=ie^{i\theta}d\theta$ gives
$\int Log(e^{i\theta})ie^{i\theta}d\theta = \int -\theta e^{i \theta}d \theta$
and it follows from there, but I am not sure if it is the same approach for $Log(z+3)$ or for a circle of radius 2. So far I am thinking $z=2e^{i \theta}$ since the radius is 2 but that's as far as I got. Do I take the same approach? Computing 
$\int_{C}Log(2e^{i \theta}+3)2ie^{i \theta}d \theta$ 
if so I have no idea where to go from there.
Thanks :)
 A: Let's apply Proposition 2 to:


*

*the circle $\gamma\colon [0,2\pi]\to \mathbb C, t\mapsto 2\exp it$,

*function $f(z)=\mathrm{Log}(z+3)$ that is continuous on a neighborhood of the circle $\mathrm{im} \,\gamma$,

*biholomorphic function $\phi(z)=z-3$.


$$\int_\gamma \mathrm{Log}(z+3)\,dz=\int_{\phi^{-1}\circ \gamma} \mathrm{Log}(z)\, dz$$
But $\phi^{-1}\circ \gamma$ is basically the circle of radius $2$ centered at $3$ and $\mathrm{Log}$ is holomorphic there.
A: You can use the same method, but from there you're gonna need to split your interval into injective portions of a substitution and then calculate it by substitution of variables noting that the exponential function is holomorphic.
Since you're computing a contour integral of a "well behaved" function you're just gonna end up at $0$. (Cauchy's Integral Theorem), since the derivative $\frac{d}{dz}Log(z+3) = \frac{1}{z+3}$ with only pole at $z=-3$ which is not on the region delimited by your contour.
If you're a math student I suggest reading a bit on substitution of variables for complex integrals and Cauchy's Integral Theorem.
