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I'm trying to use the Estimation Lemma to show that $$\left\lvert \int_{\gamma(1;2)} \frac{1}{z} \, dz \right\rvert \leq 4\pi$$ where $\gamma(1;2)$ is the circle of radius $2$ centered at $1$.

I know that the Estimation Lemma says that $$\left\lvert \int_{\gamma} f(z) \, dz \right\rvert \leq Ml$$ where $f(z)$ is such that $|f(z)|\leq M$ and $l=\int_{\alpha}^\beta |\gamma '(t)| \, dt$.

However, I'm not given a path $\gamma(t)$ so I don't understand how to apply this here. Is there an obvious path to use?

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  • $\begingroup$ What do you mean you're not given a path? Isn't $\;\gamma(1;2)\;$ the circle of radius $\;2\;$ around $\;1\;$ ? If not, what then does that mean? $\endgroup$
    – DonAntonio
    Nov 26, 2016 at 14:15
  • $\begingroup$ What does $\gamma(1;2)$ denote? $\endgroup$ Nov 26, 2016 at 14:15
  • $\begingroup$ @DonAntonio I misunderstood the notation, I've made an edit to my post making this clear now. $\endgroup$
    – MHW
    Nov 26, 2016 at 14:34

1 Answer 1

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Hint. Since $\gamma(1;2)$ is the circle centred at $1$ of radius $2$, it follows that its length is $l=2\pi\cdot 2=4\pi$.

What is an estimate for $|f(z)|=\frac{1}{|z|}$ along this circle?

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  • $\begingroup$ Oh it was the notation $\gamma (1;2)$ that was confusing me. $|f(z)| \leq 1$? $\endgroup$
    – MHW
    Nov 26, 2016 at 14:24
  • $\begingroup$ Yes, you are right. Do you mind to edit your question, and say what $\gamma(1;2)$ is? $\endgroup$
    – Robert Z
    Nov 26, 2016 at 14:28
  • $\begingroup$ I've made an edit. $\endgroup$
    – MHW
    Nov 26, 2016 at 14:33
  • $\begingroup$ @user332597 Perfect. Now your question is clear. $\endgroup$
    – Robert Z
    Nov 26, 2016 at 14:35

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