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Consider holomoprhic function $f:\mathbb C \backslash \{0\}\rightarrow \mathbb C$. I want to prove that the square:

$\partial R=[1+i,-1+i]+[-1+i,]+[-1-i,1-i]+[1-i,1+i]$

And the triangle: $\partial\Delta=[1-i,i]+[i,-1-i]+[-1-i,1-i]$

have the same line integral: $\int_{\partial R}f(z)dz=\int_{\partial \Delta}f(z)dz$

Any tips what to use here? Is the reasoning that the set is open, therefore the line integral of the triangle is zero, and therefore the line integral of the square composed of two triangles is zero? I feel that something is wrong with the answer...

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  • $\begingroup$ The two paths are homotopic in $\mathbb C\setminus \{0\}$. $\endgroup$ – Kavi Rama Murthy Nov 14 '18 at 7:48
  • $\begingroup$ You just have to show their winding numbers about the origin are the same. $\endgroup$ – Lord Shark the Unknown Nov 14 '18 at 7:48

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