Find the fallacy in the argument leading to the result $2\pi i=0$

Question

If $C$ is a positively oriented circle$ |z|=1$then by cauchy integral formula $\int_C \frac{dz}{z} =2\pi i$

Decompose$ C$ in the form $C=C_1+C_2$, Where $C_1$ is the upper half of C from $1$ to $ -1$ and $C_2$ is lower half of$ C$ from$ - 1$ to $1$, then consider the following argument and find the fallacy which leads one to conclude that $0=2\pi i$.

$\int_C \frac{dz}{z}=\int_{C_1} \frac{dz}{z}+ \int_{C_2} \frac{dz}{z}= \int_{1}^{-1} \frac{dz}{z}+ \int_{-1}^{1} \frac{dz}{z}$= $Log (z) |_{1} ^{-1}$+$Log (z) |_{-1} ^{1}$=$Log(-1)-Log(1)+Log(1)-Log(-1)=0$

I have tried so hard to find the fallacy but couldn't ... So please help me. Thanks in advance.

You have to choose a determination of Log which is a multiform function, and if I remember correctly, a crucial hypothesis in Cauchy's formula (or the residue theorem? That was long ago) is to integrate along a simple closed curve. — Sylvain Julien, May 14 '20 at 16:03

score 1 · Answer 1 · answered May 14 '20 at 16:06

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$1/z$ has a pole in $z=0$ you can't use fundamental theorem of calculus like that. Also, what is $Log(-1)$ for you?.

This actually happens to be the idea, that you have to move around the branchs of Log and that's why you have this "disphase" of $2\pi i$

answered May 14 '20 at 16:06

Julian Mejia

4,438

But the fundamental theorem is used after decomposing C into upper and lower half using $\int_{C+K} f(z) dz=$int_C f(z) dz +int_K f(z) dz – Abdullah O. Alfaqir May 14 '20 at 16:37
Each of the upper and lower parts of C could be subset of a simply connected region on which the fundamental theorem works – Abdullah O. Alfaqir May 14 '20 at 16:46
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Okay, but for each simply connected component your $Log$ is different. – Julian Mejia May 14 '20 at 17:24

score 0 · Accepted Answer · answered May 14 '20 at 16:37

When you decomposed it to $C_{1}$ and $C_{2}$ your integral is still pretty simple to solve:

All complex numbers on your countour $C_{1}$ can be written as $z=re^{i\theta}$, where $r=1$ and $0\leq\theta\leq\pi\implies z = e^{i\theta}$ on this contour. Futhermore, $dz= ie^{i\theta}d\theta$

Now, $\int_{C_1} \frac{dz}{z}= \int_{0}^{\pi} \frac{ie^{i\theta}d\theta}{e^{i\theta}}=i\int_{0}^{\pi}d\theta =i(\pi-0)=\pi i$

Similary for $C_{2}$, we have that $z=e^{i\theta}$ on this contour, where $\pi\leq\theta\leq2\pi$

Our integral on this contour becomes: $\int_{C_2} \frac{dz}{z}= \int_{\pi}^{2\pi} \frac{ie^{i\theta}d\theta}{e^{i\theta}}=i\int_{\pi}^{2\pi}d\theta =i(2\pi-\pi)=\pi i$

Add them both up to get $2\pi i$, which is consistent with your first method!

As for your method, the $Log$ function is usually defined to have a branch cut on the negative real axis so it can be single valued. I'm not sure how you've defined it.

I know about this but I'm looking for the fallacy in the argument not how to calculate the integral — Abdullah O. Alfaqir, May 14 '20 at 16:41
How do you justify using the FTOC? Isn't the single valued antiderivative function $F$, which is $=Log$ in your case, supposed to be continuous or holomorphic at all points of the contour, and depending on how you defined it, I don't think this is the case for $F=Log$, which is why I think you using it as an antiderivative function is wrong in the calculation you made. — Syed Bukhari, May 14 '20 at 16:52

Find the fallacy in the argument leading to the result $2\pi i=0$

2 Answers2