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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

Overclock
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    You could use the Cauchy-Riemann Equations to show that $\log(z-4)$ is analytic in and on $|z|=1$. Alternatively, find the anti-derivative of $\log(z-4)$ and note that it is unambiguously defined on $C$. – Mark Viola Mar 27 '18 at 19:46
  • @MarkViola This is actually a really interesting idea that didn't cross my mind at all. Any idea how (partial) derivatives of the complex logarithm differ from the standard logarithm, if at all? Cheers! – Overclock Mar 27 '18 at 19:51
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    You're welcome. My pleasure. The derivative of $\log(z)$ is $\frac1z$ on the domain of the complex logarithm (i.e., the cut plane). If you are referring to partial derivatives on the real and imaginary parts, find those parts and proceed normally. – Mark Viola Mar 27 '18 at 19:54
  • Ah wow, that simple! Brain fart on my end, I suppose. If you want to turn this into an answer I'm happy to accept it as it has certainly solved my problem! – Overclock Mar 27 '18 at 19:55

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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

$$\oint_{|z|=1} f(z)\,dz=0$$

Mark Viola
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