modulus of a line integral bounded from above

**davismj** · February 6th 2010, 06:03 PM

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did i do this one correctly?

**mabruka** · February 6th 2010, 08:25 PM

Well i would use something like:

$\left| \int_{\delta R} \frac{u(z)}{(z-z_0)^2} dz \right| \leq CR\int_{0}^{2\pi} \left| \frac{1}{(Re^{it}-z_0)^2} dt \right| \leq \frac{2\pi CR}{||R|-|z_0||^2}$

the last part goes to 0 as R grows

What do u think?

I think there is something wrong with your attempt, your bound must not depend on t.

**davismj** · February 6th 2010, 09:28 PM

Originally Posted by mabruka

Well i would use something like:

$\left| \int_{\delta R} \frac{u(z)}{(z-z_0)^2} dz \right| \leq CR\int_{0}^{2\pi} \left| \frac{1}{(Re^{it}-z_0)^2} dt \right| \leq \frac{2\pi CR}{||R|-|z_0||^2}$

the last part goes to 0 as R grows

What do u think?

I think there is something wrong with your attempt, your bound must not depend on t.

I agree. Can you explain, though, why $(Re^{it}-z_0)^2 = ||R|-|z_0||^2$ is true?

**mabruka** · February 6th 2010, 10:17 PM

Well i thouht using triangle inequality and then just squaring it up:

$||Re^{it}|-|z_0||\leq | Re^{it}- z_0| \Rightarrow \frac{1}{| Re^{it}- z_0| } \leq \frac{1}{||Re^{it}|-|z_0||}=\frac{1}{||R|-|z_0||}$

where obviously $|Re^{it}|=|R|=R$

What do you think?

**davismj** · February 7th 2010, 12:39 PM

Ah yes, you're absolutely right. Thank you so much, this makes perfect sense to me now.

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