Find the complex integral

Question

I have $$\int_{\gamma} {\frac{z}{\overline z}}dz$$ where $\gamma$ is the edge of $\{1 < |z| < 2\ $and $\Im (z) > 0\}$.

I think the way to solve this is to calculate the integral for $|z|=1$ and then $|z|=2$, then subtract the first result from the second one.

For $|z|=1$, the integral is $i\pi$. For $|z|=1$, the integral is $4i$. So my answer is $4i-i\pi$. But the answer in book is ${\frac{4}{3}}$ .

Note that $\frac{z}{\bar{z}} =\frac{z^2}{|z|^2}$. – CyclotomicField Nov 29 '18 at 18:51 — CyclotomicField, Nov 29 '18 at 18:51

José Carlos Santos · Accepted Answer · 2018-11-29T19:21:41.543

0

Actually, in the upper semi-circle $\lvert z\rvert=1$, you have$$\frac z{\overline z}=\frac{z^2}{\lvert z\rvert^2}=z^2$$and therefore the integral (if you move along the semi-circle from the left to the right) is equal to $\dfrac23$. By the same argument, the integral along the larger semicircle (this time moving from the right to left) is $-\dfrac43$.

Can you take it from here?

edited Nov 29 '18 at 19:21

answered Nov 29 '18 at 18:55

José Carlos Santos

427,504

I don't understand how you got the 2/3 and -1/6. Is this the domain mentioned in question? https://ibb.co/RYY7qBp – user3132457 Nov 29 '18 at 19:06
I used the fact that a primitive of $z^2$ is $\frac{z^3}3$. So, the first of those integrals is $\frac{1^3}3-\frac{(-1)^3}3=\frac23$. The other one is similar. – José Carlos Santos Nov 29 '18 at 19:08
For |z|=2, I get $$\int {\frac{z}{\overline z}}dz = \int {\frac{z^2}{|z|^2}} = \int {\frac{z^2}{4}} = {\frac{z^3}{3}} \times {\frac{1}{4}} = {\frac{1}{12}}(8+8) = {\frac{16}{12}} = {\frac{4}{3}}$$ which looks like the final answer. – user3132457 Nov 29 '18 at 19:19
@user3132457 Yes, I made a mistake. I've edited my answer. – José Carlos Santos Nov 29 '18 at 19:22
Is it -4/3 or 4/3? – user3132457 Nov 30 '18 at 05:18
It is $-\frac43$. When we are doing a contour integration, you must be moving along that contour in a certain direction. I assumed that in the case of the larger semicircle you were moving from the right to the left. Otherwise, yes, the answer would be $\frac43$. – José Carlos Santos Nov 30 '18 at 07:19
Let us continue this discussion in chat. – user3132457 Nov 30 '18 at 10:35
How do I calculate $\int_{-2}^1{\frac{z}{\overline z}}dz$? – user3132457 Jan 23 '19 at 18:59
It's the part of the solution, I calculated the integrals over circles, but can't calculate for the (-2, -1) and (1, 2) lines – user3132457 Jan 23 '19 at 19:04

Find the complex integral

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