Question-134596

Question Number 134596 by mohammad17 last updated on 05/Mar/21
Commented by mohammad17 last updated on 05/Mar/21
$${help}\:{me}\:{sir} \\$$
Answered by Olaf last updated on 05/Mar/21
$$\left.{a}\right) \\$$$$\int_{\mathrm{C}} \left({z}+\mathrm{1}\right){dz}\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \left({x}+\mathrm{1}+{ix}^{\mathrm{2}} \right){dx} \\$$$$=\:\left[\frac{{x}^{\mathrm{2}} }{\mathrm{2}}+{x}+{i}\frac{{x}^{\mathrm{3}} }{\mathrm{3}}\right]_{\mathrm{0}} ^{\mathrm{1}} \:=\:\frac{\mathrm{3}}{\mathrm{2}}+\frac{{i}}{\mathrm{3}} \\$$$$\left.{b}\right) \\$$$$\int_{\mathrm{C}} \left({x}^{\mathrm{2}} +{iy}^{\mathrm{3}} \right){dz} \\$$$$=\:\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}} \left({r}\mathrm{cos}^{\mathrm{2}} \frac{\pi}{\mathrm{4}}+{ir}^{\mathrm{3}} \mathrm{cos}^{\mathrm{3}} \frac{\pi}{\mathrm{4}}\right){dr} \\$$$$=\:\left[\frac{{r}^{\mathrm{2}} }{\mathrm{4}}+{i}\frac{{r}^{\mathrm{4}} }{\mathrm{8}\sqrt{\mathrm{2}}}\right]_{\mathrm{0}} ^{\sqrt{\mathrm{2}}} \:=\:\frac{\mathrm{1}}{\mathrm{2}}+\frac{{i}}{\:\mathrm{2}\sqrt{\mathrm{2}}} \\$$$$\left.{c}\right) \\$$$$\int_{\mathrm{C}} {xdz}\:=\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} {r}\mathrm{cos}\theta{d}\theta\:=\:\mathrm{0} \\$$
Commented by mohammad17 last updated on 05/Mar/21
$${sir}\:{why}\:\left({x}+{ix}^{\mathrm{2}} \right){replace}\:{z}\:{in}\:\left({a}\right) \\$$
Commented by Olaf last updated on 05/Mar/21
$$\mathrm{because}\:{z}\:=\:{x}+{iy}\:\mathrm{and}\:{y}\:=\:{x}^{\mathrm{2}} \\$$
Commented by mohammad17 last updated on 05/Mar/21
$${yes}\:{yes}\:{sir}\:{thank}\:{you}\: \\$$