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Question Number 136682 by snipers237 last updated on 24/Mar/21

$$Im\left({\int }_{C^{+}(0,\frac{1}{2})}\stackrel{-}{zdz}\right)=\frac{\pi }{2}$$

Answered by Olaf last updated on 24/Mar/21

$$\mathrm{\Omega }=\mathrm{Im}{\int }_{{\mathrm{C}}^{+}(0,\frac{1}{2})}\stackrel{-}{zdz}$$$$\mathrm{\Omega }=\mathrm{Im}{\int }_0^{2\pi }\frac{1}{2}{e}^{-i\theta }d\left(\frac{1}{2}{e}^{i\theta }\right)$$$$\mathrm{\Omega }=\frac{1}{4}\mathrm{Im}{\int }_0^{2\pi }id\theta$$$$\mathrm{\Omega }=\frac{1}{4}\mathrm{Im}\left(2\pi i\right)=\frac{2\pi }{4}=\frac{\pi }{2}$$

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