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Hi guys, I've calculated the first part, but I don't understand how it is related with triangle inequality. Can someone help?Thanks!

  • $\begingroup$ In this context, the triangle inequality says that the modulus of the first integral is less than or equal to the second integral. $\endgroup$ – Greg Martin Sep 18 '16 at 7:56

First, parametrize the $\{|z| = 1\}$ by $z = e^{i\theta}$. Then the first integral becomes \begin{align} \int_{|z|=1} (z-1)\ dz = \int^{2\pi}_0 (e^{i\theta}-1)ie^{i\theta}\ d\theta = 0. \end{align} For the second integral, we have \begin{align} \int_{|z|=1} |z-1||dz| = \int^{2\pi}_0 |e^{i\theta}-1| dt = \int^{2\pi}_0 |\cos\theta-1 + i\sin\theta|\ dt = \int^{2\pi}_0 \sqrt{(1-\cos\theta)^2+\sin^2\theta}\ d\theta = \int^{2\pi}_0\sqrt{2-2\cos\theta}\ d\theta = 2\int^{2\pi}_0 |\sin(\theta/2)|\ d\theta = 8. \end{align} Indeed, we have the triangle inequality \begin{align} \left|\int_{|z|=1}(z-1)\ dz\right| \leq \int_{|z|=1} |z-1||dz|. \end{align}

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