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In an homework exercise, we're asked to integrate $\int_{C} \frac{1}{k-\bar{z}}dz$ where C is some circle that doesn't pass through $k$.

I tried solving this question through two different approaches, but have arrived at different answers.

Idea: use the fact that in C, $z\bar{z}=r^2$, where $r$ is the radius of C, to get

$$\int_{C} \frac{1}{k-\frac{r^2}{z}}dz$$

We then apply Cauchy to get that the answer is $2πi r^2/k^2$ when C contains $r^2/k$, and 0 otherwise.

Another idea: Intuitively, since C is a circle we get ${\int_{C} \frac{1}{k-\bar{z}}dz} = \int_{C} \frac{1}{k-z}dz$ (since $\bar{z}$ belongs to C iff $z$ belongs to C) and we can then use Cauchy's theorem and Cauchy's formula (depending on what C contains) to arrive at a different answer.

Which answer (if any) is correct?

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is the circle centered at the origin? (if so, the first answer looks legit to me) –  uncookedfalcon Jan 13 '13 at 8:33
    
I hadn't noticed this until now (which brings up the question of how I actually arrived at any of this, to which I have no answer), but yeah, the question defines it at the origin. –  ta33 Jan 13 '13 at 8:36
    
to see the second thing isn't true, take $k = 0$, and $r = 1$, $\int_C \frac{1}{\bar{z}} = \int_C z = 0$, $\int_C \frac{1}{z} = 2\pi i$ –  uncookedfalcon Jan 13 '13 at 8:41
    
I had thought about this myself, but I don't understand why it isn't true, and whether the first explanation is... (I'm heading out for a little bit, so I won't be here to respond) –  ta33 Jan 13 '13 at 8:43

3 Answers 3

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I've definitely made the same "intuitive" argument about $z \in C$ iff $\bar{z} \in C$, so eh, the integrals are probably the same. In retrospect, this is because I didn't understand, formally or intuitively, what $dz$ means.

First off, what is a contour integral? For concreteness, let's work with a circle $C \subset \mathbb{C}$ and a (continuous) function $f: C \rightarrow \mathbb{C}$. This is literally all you do: wind around the circle (counterclockwise) and along the way, pick out some points $z_0, \ldots, z_n$. Just as in integrating functions $f: \mathbb{R} \rightarrow \mathbb{R}$, you consider the quantity $\sum_i f(z_i)(z_i - z_{i-1})$:enter image description here

For roughly the same reason as its true in real analysis, when you take finer and finer samplings, these all approach a common value, which we call $\int_C f dz$.

Now, from this it should be clear heuristically that when $C$ is centered at the origin, $\int_C zdz$ need not equal $\int_C \bar{z} dz$, since when you exchange $z_i$ and $\bar{z_i}$, you'd also need to exchange $(z_i - z_{i-1})$ with $(\bar{z}_i - \bar{z}_{i-1})$ (with a sign to account for switching from counterclockwise to clockwise) but you don't when you leave the $dz$ untouched.

P.S. For fun, consider working out that $\int_{|z| = 1} zdz = 0, \int_{|z|=1} \bar{z} dz = 2 \pi i$ straight from this definition. I tried sampling at the $n^{th}$ roots of unity, for $zdz$ I got 0, and for $\bar{z}dz$ I got $n(1 - \zeta^{-1})$, where $\zeta = e^{2\pi i/n}$.

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I think that the first approach is correct. If the circle is centered about the origin, then it is true that the conjugate also lies on the same circle. However, "dz" at $ z $ is not the same as "dz" at $ \bar{z} $ . This is why they are different.

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The original function is not holomorphic since $\frac{d}{d\overline{z}}\frac{1}{k-\overline{z}}=\frac{-1}{(k-\overline{z})^{2}}$. So you cannot apply Cauchy's integral formula.

Let $C$ be centered at $c$ with radius $r$, then we have $z=re^{i\theta}+c=c+r\cos(\theta)+ri\sin(\theta)$, and its conjugate become $c+r\cos(\theta)-ri\sin(\theta)=c+re^{-i\theta}$. Therefore we have $k-\overline{z}=k-c-re^{-i\theta}$. To noramlize it we mutiply $(k-c-r\cos(\theta))-ri\sin(\theta)=k-c-re^{i\theta}$. The result is $(k-c-r\cos(\theta))^{2}+r^{2}\sin^{2}\theta$. Rearranging gives $$k^{2}+c^{2}+r^{2}-2kc-2(k-c)r\cos(\theta)=A-2Br\cos(\theta),A=k^{2}+c^{2}+r^{2}-2kc,B=k-c$$

So we have $$\int_{C}\frac{1}{k-\overline{z}}=\int^{2\pi}_{0}\frac{k-c-re^{i\theta}}{(k-c-re^{-i\theta})(k-c-re^{-i\theta})}dre^{i\theta}=ri\int^{2\pi}_{0}\frac{Be^{i\theta}-re^{2i\theta}}{A-2Br\cos(\theta)}d\theta$$

So it suffice to intergrate $$\frac{e^{i\theta}}{1-C\cos(\theta)}d\theta,C=constant$$ since the above integral is a linear combination of integrals of this type. But this should be possible in general as we can make trignometric substitutions.

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I agree that $\frac{1}{k - \bar{z}}$ isn't meromorphic. I think his argument is that on the circle $C$ centered at 0 with radius $r$, we have this function agrees with $\frac{1}{k - \frac{r^2}{z}}$, which is meromorphic, and so we can apply Cauchy to that (the point being that the contour integral only "sees"/"cares about" the values on the circle) –  uncookedfalcon Jan 13 '13 at 10:55
    
The circle is not necessarily having 0 as the origin, though. –  Bombyx mori Jan 13 '13 at 11:46
    
OP mentioned that it was in the comments :(... –  uncookedfalcon Jan 13 '13 at 11:48
    
I see, then it is much simpler. –  Bombyx mori Jan 13 '13 at 11:51

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