Question Number 214340 by issac last updated on 06/Dec/24
$$\mathrm{evaluate} \\ $$$$\frac{\oint_{{C}} \:\frac{{z}}{\mathrm{2}\centerdot\mathrm{sin}\left({z}\right)−\mathrm{2}{z}\centerdot\mathrm{cos}\left({z}\right)−\pi}\mathrm{d}{z}}{\oint_{\:{C}} \:\frac{\mathrm{2}}{\mathrm{2}\centerdot\mathrm{sin}\left({z}\right)−\mathrm{2}{z}\centerdot\mathrm{cos}\left({z}\right)−\pi}\mathrm{d}{z}} \\ $$$$\mathrm{where}\:{C}\:\mathrm{is}\:\mathrm{the}\:\mathrm{circle}\mid{z}−\frac{\mathrm{3}\pi}{\mathrm{4}}\mid=\frac{\pi}{\mathrm{4}}. \\ $$
Answered by MrGaster last updated on 24/Dec/24
$$\oint_{{C}} \frac{{z}}{\cancel{\mathrm{2}\centerdot\mathrm{sin}\left({z}\right)−\mathrm{2}{z}\centerdot\mathrm{cos}\left({z}\right)−\pi}}{dz} \\ $$$$\oint_{{C}} \frac{\mathrm{2}}{\cancel{\mathrm{2}\centerdot\mathrm{sin}\left({z}\right)−\mathrm{2}{z}\centerdot\mathrm{cos}\left({z}\right)−\pi}}{dz} \\ $$$$=\frac{\oint_{{C}} {z}\:{dz}}{\oint_{{C}} \mathrm{1}\:{dz}} \\ $$$$\underset{=\frac{{Cauchy}^{,} {s}\:{inegral}\:{theorem}}{\mathrm{2}\pi{i}\centerdot\mathbb{Z}}} {\underbrace{\oint_{{C}} {z}\:{dz}=\mathrm{0}}} \\ $$$$=\frac{\mathrm{0}}{\mathrm{2}\pi{i}\centerdot\mathbb{Z}} \\ $$$$=\mathrm{0} \\ $$