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I am trying to find the integral

$$\oint_c Re(z)\;dz$$ where c is a circle $$|z|=2$$

I don't know what to do.

I tried some things but I don't know if I am correct. $e^{i\theta} = \cos \theta +i \sin \theta$, so $Re(z) = \cos \theta$? And then $$\int_0^{2\pi} 2\cos \theta i e^{i\theta}\;d\theta$$

Can someone explain a little what's going on cause I am trying to understand it 2 hours now and I am getting nervous.

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1 Answer

up vote 2 down vote accepted

Hint: This is the correct approach, you are missing a factor of $2$. This is because on the circle $|z|=2$ we have can parametrize by setting $z=2e^{i\theta}$, and then $\text{Re}(z)=2\cos \theta$ and $dz=2ie^{i\theta}.$ The integral is $$4i\int_0^{2\pi} \cos(\theta)e^{i\theta}d\theta,$$ and we can rewrite $\int_0^{2\pi} \cos(\theta)e^{i\theta}d\theta$ as $$\int_0^{2\pi} \cos^2(\theta)d\theta+i\int_0^{2\pi} \cos(\theta)\sin(\theta)d\theta$$ by using $e^{i\theta}=\cos(\theta)+i\sin(\theta)$. The second integral is zero by symmetry, and the first is equal to $\pi$. To see why it equals $\pi$, notice that by symmetry it is $\int_0^{2\pi}\sin^2(\theta)d\theta$, so that adding them together gives twice the integral which is $$\int_0^{2\pi}\sin^2(\theta)+\cos^2(\theta)d\theta=\int_0^{2\pi}d\theta=2\pi.$$ Usual substitutions would of also worked. Thus you get $4\pi i$

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Thanks. I knew I was close. –  tripons Jan 26 '12 at 17:41
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