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Goursat's theorem: If $\Omega$ is an open set in $\mathbb C,$ and $T\subset \Omega$ a triangle whose interior is also contained in $\Omega,$ then: $$\int_{T}f(z)=0$$ whenever f is holomorphic in $\Omega.$ here is the prove image from Complex Analysis by Stein and Shakarchi.

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There is another exercise relate to this:

If $\Omega$ is an open set in $\mathbb C,$ and $T\subset \Omega$ a triangle whose interior is also contained in $\Omega,$ Suppose $f$ is a function holomorphic in $\Omega$, except possibly at a point $w$ inside T, if $f$ is bounded near $w$, then: $$\int_{T}f(z)=0$$ My question: what method do we use to solve the problem here and when a question of exception like this is been asked?

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  • $\begingroup$ Since $f(z)$ is bounded near $w$ it has a removable discontinuity at $w$ . $\endgroup$ Nov 11, 2019 at 7:32
  • $\begingroup$ What I was think is that f(w) can be replaced by $f(z_0)+f'(z_0)(w-z_0)+\psi(z)(w-z_0).$ $\endgroup$
    – John He
    Nov 11, 2019 at 7:38
  • $\begingroup$ I would just use the fact that $f$ can be defined to be holomorphic on a neighborhood of $T$ and so the function satisfies the conditions of the theorem you shared. $\endgroup$ Nov 11, 2019 at 7:42
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    $\begingroup$ You could draw a very small triangle around $w$, then connect its vertices to the vertices of the original triangle. This gives you a few triangles not containing $w$ to which you can apply Goursat's theorem. It follows easily that the integral along the very small triangle is equal to the integral along the initial triangle. But the integral along the small triangle is very close to $0$ because $f$ is bounded. $\endgroup$ Nov 11, 2019 at 7:43
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    $\begingroup$ Or you could have a small keyhole inside the triangle. The integral over all the lines of the outer triangle should go to zero, because they are in a region that does not include the point w. Over a small neighborhood of the circle, the absolute value of the integral is proportional to the radius of the small neighborhood. For a small enough neighborhood, this goes to zero, so the whole integral goes to zero.? @Gribouillis does this sound right? $\endgroup$
    – u_any_45
    Sep 21, 2020 at 16:34

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