Question Number 220790 by SdC355 last updated on 19/May/25
$$\mathrm{Complex}\:\mathrm{integral} \\ $$$$\oint_{\:\mathrm{C}} \:\frac{\mathrm{d}{z}}{{z}^{\mathrm{3}} +\mathrm{1}}=??\:,\:{C};{x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{4} \\ $$$$\oint_{\:{C}} \:\frac{\mathrm{1}}{{z}}{e}^{{z}} \:\mathrm{d}{z},\:{C};\begin{cases}{{y}=\mathrm{1}\:,\:−\mathrm{1}\leq{x}\leq\mathrm{1}}\\{{y}=−\mathrm{1}\:,\:−\mathrm{1}\leq{x}\leq\mathrm{1}}\\{{x}=\mathrm{1}\:,\:−\mathrm{1}\leq{y}\leq\mathrm{1}}\\{{x}=−\mathrm{1}\:,\:−\mathrm{1}\leq{y}\leq\mathrm{1}}\end{cases}\: \\ $$
Answered by vnm last updated on 19/May/25
$$\frac{\mathrm{1}}{{z}^{\mathrm{3}} +\mathrm{1}}=\frac{{a}}{{z}+\mathrm{1}}+\frac{{b}}{{z}−{e}^{{i}\frac{\pi}{\mathrm{3}}} }+\frac{{c}}{{z}−{e}^{−{i}\frac{\pi}{\mathrm{3}}} }= \\ $$$$\frac{\left({a}+{b}+{c}\right){z}^{\mathrm{2}} +\left(..\right){z}+\left(..\right)}{{z}^{\mathrm{3}} +\mathrm{1}}\Rightarrow{a}+{b}+{c}=\mathrm{0} \\ $$$$\oint_{\mid{z}\mid=\mathrm{2}} \frac{{dz}}{{z}^{\mathrm{3}} +\mathrm{1}}={a}\oint_{\mid{z}\mid=\mathrm{2}} \frac{{dz}}{{z}+\mathrm{1}}+{b}\oint_{\mid{z}\mid=\mathrm{2}} \frac{{dz}}{{z}−{e}^{{i}\frac{\pi}{\mathrm{3}}} }+{c}\oint_{\mid{z}\mid=\mathrm{2}} \frac{{dz}}{{z}−{e}^{−{i}\frac{\pi}{\mathrm{3}}} }= \\ $$$${a}\mathrm{2}\pi{i}+{b}\mathrm{2}\pi{i}+{c}\mathrm{2}\pi{i}=\left({a}+{b}+{c}\right)\mathrm{2}\pi{i}=\mathrm{0} \\ $$$$\oint_{{C}} \frac{{e}^{{z}} }{{z}}{dz}=\oint_{{C}} \frac{{e}^{{z}} }{{z}−\mathrm{0}}=\pm{e}^{\mathrm{0}} \mathrm{2}\pi{i}=\pm\mathrm{2}\pi{i} \\ $$$$\mathrm{The}\:\mathrm{sign}\:\mathrm{depends}\:\mathrm{on}\:\mathrm{whethere} \\ $$$$\mathrm{the}\:\mathrm{contour}\:\mathrm{is}\:\mathrm{traversed} \\ $$$$\mathrm{clockwise}\left(−\right)\:\mathrm{or}\:\mathrm{anticlockwise}\left(+\right) \\ $$
Commented by SdC355 last updated on 19/May/25
$$\mathrm{thx} \\ $$