Regarding Cauchy - Goursat theorem with Log function If $f(z) = \operatorname{Log}{z+2} $ and $C$ is $|z|=1$ , can Cauchy-Goursat theorem be applied at all?
I was having the impression that log function resemble a ray $$\ln{r}+i\theta$$ therefore there is no simple closed curve possible to enclosed the log function so that it is analytic?
But it seems like $\int_C f(z)=0$
Why is that
 A: As you say, Cauchy's integral theorem is not applicable since $f$ is not holomorphic on $C = \{ |z| = 1 \}$. I'm assuming that you mean the principal branch of $\log$.
However, the integral can still be computed in several different ways. You need to interpret
$$
\int_C (\log z + 2) \,dz = \lim_{\varepsilon\to 0^+} \int_{C_\varepsilon} (\log z + 2)\,dz,
$$ 
where $C_\varepsilon$ is the curve $z = e^{it}$, $-\pi+\varepsilon \le t \le \pi-\varepsilon$. Then $f$ is holomorphic on a neighbourhood of $C_\varepsilon$ and the integral over $C_\varepsilon$ can be computed using the (complex version) of the fundamental theorem of calculus. An anti-derivative of $\log z + 2$ is $z(1+\log z)$, so
\begin{align}
\int_{C_\varepsilon} (\log z + 2)\,dz &= \left[ z(1+\log z)  \right]^{\exp(i(\pi-\varepsilon))}_{\exp(i(-\pi+\varepsilon))} \\
&=  \exp(i(\pi-\varepsilon)(1+i(\pi-\varepsilon)) - \exp(-i(\pi-\varepsilon)(1-i(\pi-\varepsilon)) \\
&\to \exp(i\pi)(1+i\pi) - \exp(-i\pi)(1-i\pi) = -1(1+i\pi) + (1-i\pi) = -2\pi i
\end{align}
as $\varepsilon \to 0$.
You can also compute the integral by direct parametrization, or by integrating along an indented keyhole contour avoiding the branch cut of $\log$.
A: No, you can't, $f$ is not even defined on $C$.
On the other hand if you meant $f(z) = \mathrm{Log}(z+2)$, then sure.
A: If you really mean $f(z)=\text{Log}(z+2)$, then you can use the CG-theorem. Of course, the function $\text{Log}(z)$ has a branch cut along the negative real axis. But if you look at the function $\text{Log}(z+2)$ the branch cut starts at $z=-2$, so $\text{Log}(z+2)$ is analytic in the disc with radius one and you can use the CG-therom to show that the integral is zero.
Greets,
Satonka
