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Let $\gamma$ be a close piecewise smooth contour consisting of the three straight lines from $1$ to $2$, from $2$ to $1+i$ and from $1+i$ to $1$.

I have to show that $$\left|\int_{\gamma}\frac{1}{\overline z +i}\,dz\right|\leq 2+\sqrt{2}$$

I know that I can use the estimation lemma to bound $$\left|\frac{1}{\overline z +i}\right|\leq M$$ and I think that must be $M=1$, because $\lvert\gamma\rvert=2+\sqrt2$, but I can't find a suitable upper bound. Any hint?

Maybe $$\left|\frac{1}{\overline z +i}\right|= \frac{1}{\lvert\overline z +i\rvert}\leq\frac{1}{\lvert\overline z\rvert - \lvert i\rvert}=\frac{1}{\lvert z\rvert - 1}$$ but this does not give me any further information since $1\leq\lvert z\rvert\leq 2$

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  • $\begingroup$ Use $|A| = |\bar A|$. $\endgroup$
    – aschepler
    Feb 23, 2023 at 0:30
  • $\begingroup$ @aschepler do you mean something like $\left|\overline{\frac{1}{\overline z + i}}\right|=\left|\frac{1}{z- i}\right|$...but I don't see any boundary again $\endgroup$
    – James Cats
    Feb 23, 2023 at 0:40

1 Answer 1

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Find the points where $\left|\frac1{\overline z+i}\right|\le1$: $$|\overline z+i|\le1\iff\overline z\text{ is in unit circle centred on }-i$$ $$\iff z\text{ is in unit circle centred on }i$$ Since all parts of the curve lie on or outside the unit circle centred on $i$, $\left|\frac1{\overline z+i}\right|\le1$ on all of $\gamma$ and we get the desired result.

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  • $\begingroup$ Very intuitive way, thanks! $\endgroup$
    – James Cats
    Feb 23, 2023 at 1:00
  • $\begingroup$ @GiacomoGatti Please upvote and accept my answer :) $\endgroup$ Feb 23, 2023 at 1:12

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