The question I have to answer is
Let $C=\{z:|z|=1\}$. Prove that $\int_{C}f(z)dz=0$ where $f(z)=Ln(z-4)$
My attempt:
Initially my first idea was to rewrite the complex logarithm in terms of the real logarithm and its argument, namely $Ln(z-4)=ln|z-4|+i(arg(z-4)+2\pi k)$ but instantly I run into the problem where this is in terms of $z$ whereas usually we are solving things in the form $ln(x+iy)$ where it is easy to find the modulus and argument.
Immediately we can see that $Ln(z-4)$ has a branch point at $z=4$ and my thought process was that if this is the case then would $Ln(z-4)$ just be analytic $\forall z \in C$ and thus, by Cauchy's Theorem that states if $f(z)$ is analytic $\forall z \in C$ then $\int_{C}f(z)dz=0$, we simply have that $\int_{|z|=1}Ln(z-1)dz=0$. I have absolutely zero concrete evidence behind this and that's exactly what my question is: how can I prove this is true.
If someone can point me along the right track I'd appreciate it. I know that I have to end up with Cauchy's Theorem telling me that $f(z)$ is analytic in $C$ but I'm not sure how to get there due to the complex logarithm (have done the other questions in the set with no problem).
Thanks
Note that if we cut the plane along the positive real axis from $z=4$ to $z=\infty$, we have
$$\log(z-4)=u(x,y)+iv(x,y)$$
where
$$u(x,y)=\frac12 \log((x-4)^2+y^2) \tag 1$$
and
$$v(x,y)=\pi+\arctan\left(\frac{y}{x-4}\right)\tag 2$$
It is easy to show that $u$ and $v$ as given by $(1)$ and $(2)$ satisfy the Cauchy-Riemann Equations when $(x,y)$ are on the closed disk $x^2+y^2=1$.
Therefore, Cauchy's Integral Theorem guarantees that
$$\oint_{|z|=1} f(z)\,dz=0$$