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A textbook I've been reading contains the question:

What is $$\oint_C\frac{z}{z-1}dz$$ where $C$ is a circle with radius $4$ centered at the origin?

Apparently, the correct answer is $2\pi i$, but I think the answer is $0$.

Here is my working:

$C$ is parametrized as $\gamma(t)=4e^{it}, t\in [0,2\pi]$. Therefore, $$\begin{align}\oint_C\frac{z}{z-1}dz &= \oint_C1+\frac{1}{z-1}dz \tag1\\[0.5em] &= \int_0^{2\pi}4ie^{it}+\frac{4ie^{it}}{4e^{it}-1}dt \tag2\\[0.5em] &= [4e^{it}]_0^{2\pi}+[\ln|4e^{it}-1|]_0^{2\pi} \tag3\\[0.5em] &=0 \tag4 \end{align}$$

Where did I mess up?

Blue
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YWu
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  • You messed up with $\ln$. There are two different branches to consider, which differ by $2\pi i$. (As a general rule in complex analysis when you first start out: if you calculation goes wrong, the first thing to check are things like square roots, cube roots, and $\log$, because these are almost always the cause of errors) – peek-a-boo Aug 04 '21 at 02:32
  • Consider instead a simpler problem: what is $\int_{\gamma}\frac{dz}{z}$, where $\gamma: t\mapsto e^{it}$ is the unit circle parametrized in the positive sense? – peek-a-boo Aug 04 '21 at 02:54
  • @peek-a-boo It makes sense in this example because you have $\int_{-pi}^\pi \frac{ie^{it}}{e^{it}}dt=\int_{-pi}^\pi i dt = 2\pi i$ – YWu Aug 04 '21 at 02:58
  • yes, so here the correct answer is $2\pi i$. But the naive (and incorrect) answer is that $\int_{\gamma}\frac{dz}{z}=\int_{\gamma}\frac{d}{dz}(\log z),dz=\log(\gamma(2\pi))-\log(\gamma(0))=0$, since $\gamma(0)=\gamma(2\pi)$. Do you see what went wrong here? The issue is that there is no holomorphic function $\log:\Bbb{C}\setminus{0}\to\Bbb{C}$ such that $\log'(z)=\frac{1}{z}$ for all $z$. i.e $z\mapsto \frac{1}{z}$ has no holomorphic primitive defined on all of $\Bbb{C}\setminus{0}$. You made precisely this error (just with a slightly different curve and integrand). – peek-a-boo Aug 04 '21 at 03:11

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As pointed out in the comments, you could use the log function, but$\textit{ with care!}$

$$\left. \log \left(4e^{it}-1\right)\right|_{0}^{2\pi}=\left. \arg \left(4e^{it}-1\right)\right|_{0}^{2\pi} =2\pi i.$$

Alternatively, let $z=1+ 4 e^{i\theta}$ so that $dz = 4i e^{i\theta}\,d\theta $

$$\oint \frac{z}{z-1} \, dz= \int_0^{2\pi} \frac{1+4e^{i\theta}}{4e^{i\theta}}\, 4i e^{i\theta}\, d\theta= i\int_0^{2\pi} d\theta + 4i \int_0^{2\pi} e^{i\theta}\, d\theta =2\pi i. $$

mjw
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