I'm having trouble understanding the difference, other than notation, between a contour integral over an open curve and a contour integral over a closed curve. So far, it seems to me that the difference is only in the limits of integration.

More specifically, I'm working on proving this: $$\lvert\oint_\Gamma \frac{cos(z)}{z}dz\rvert \le 2e\pi$$ where the path traces the unit circle once.

I know how to prove a very similar problem, just without the closed circle:

$$\lvert\int_\gamma \frac{cos(z)}{z}dz\rvert \le 2e\pi$$

To prove this I use the theorem,

$$\lvert\int_C f(z)dz\rvert \le ML$$

and then prove that

$$L= 2\pi$$ and $$M=e.$$

Can I go about this proof with the closed circle the same way?

  • $\begingroup$ You can use \left and \right to make the magnitude bars adapt to the size of their content. $\endgroup$
    – joriki
    Commented Apr 20, 2016 at 14:27

1 Answer 1


Yes. $ML$ estimation doesn't require the curve to be open (or closed), so you can certainly use the $ML$ estimate argument in the case where $\Gamma$ is the unit circle traversed once.

  • $\begingroup$ $M$ is the maximum modulus and $L$ the length of the contour ? $\endgroup$
    – reuns
    Commented Apr 20, 2016 at 16:08
  • $\begingroup$ Yes, that's right. $\endgroup$
    – user307169
    Commented Apr 20, 2016 at 16:15

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