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I have this integral to calculate:

$$I=\int_{|z|=2}(e^{\sin z}+\bar z)dz.$$

I do it this way:

$$I=\int_{|z|=2}e^{\sin z}dz+\int_{|z|=2}\bar zdz.$$

The first integral is $0$ because the function is holomorphic everywhere and it is a contour integral. As for the second one, I have

$$\int_{|z|=2}\bar zdz = \int_0^{2\pi}e^{-i\theta}\cdot 2 d\theta=-\int_0^{-2\pi}e^{i\tau}\cdot 2 d\tau=\int_0^{2\pi}e^{i\tau}\cdot 2 d\tau=\int_{|z|=2}zdz=0$$

because the function is now holomorphic.

It seems fishy to me. Is it correct?

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The integral of $\bar{z}$ over a closed curve is $2i$ times the area enclosed by the curve. It follows from Green's theorem. –  PAD Dec 16 '12 at 23:15

2 Answers 2

up vote 3 down vote accepted

If $z = 2e^{i \theta}$, then $$\bar{z} dz = 2e^{-i \theta}2i e^{i \theta} d \theta = 4i d \theta$$ Hence, $$\int_{\vert z \vert = 2} \bar{z} dz = \int_0^{2 \pi} 4i d \theta = 8 \pi i$$

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Another method for the second integral:

$$ \begin{align*} \int_{|z|=2}\bar z\,dz &= \int_{|z|=2} \frac{z \bar z}{z} \,dz \\ &= \int_{|z| = 2} \frac{|z|^2}{z}\,dz \\ &= \int_{|z| = 2} \frac{4}{z}\,dz \\ &= 4 \cdot 2\pi i \\ &= 8 \pi i. \end{align*} $$

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