Question Number 135635 by mohammad17 last updated on 14/Mar/21
Commented by mohammad17 last updated on 14/Mar/21
$${help}\:{me}\:{sir}\: \\ $$
Answered by Olaf last updated on 14/Mar/21
![Ω = ∫_C (e^z −z^_ )dz Let O(z=0), A(3i), B(−4) Ω = ∫_(OB) (e^z −z^_ )dz+∫_(BA) (e^z −z^_ )dz+∫_(AO) (e^z −z^_ )dz Ω = ∫_(OB) (e^z −z^_ )dz+∫_(BA) (e^z −z^_ )dz+∫_(AO) (e^z −z^_ )dz OB : z = x, x goes to 0 to −4 BA : z = x+iy = x +i((3/4)x+3) z = (1+((3i)/4))x+3i x goes to −4 to 0 AO : z = y, y goes to 3 to 0 Ω = ∫_0 ^(−4) (e^x −x)dx+∫_(−4) ^0 (e^(3i) e^((1+((3i)/4))x) −(1−((3i)/4))x+3i)dx +∫_3 ^0 (e^y −y)dy Ω = [e^x −(x^2 /2)]_0 ^(−4) +[(e^(3i) /(1+((3i)/4)))e^((1+((3i)/4))x) −(1/2)(1−((3i)/4))x^2 +3ix]_(−4) ^0 +[e^y −(y^2 /2)]_3 ^0 Ω = ((1/e^4 )−9)+((e^(3i) /(1+((3i)/4)))(1−e^(−4−3i) )+8(1−((3i)/4))+12i) +(((11)/2)−e^3 ) Ω = (1/e^4 )−e^3 +(9/2)+6i+(4/5)(4−3i)(e^(3i) −(1/e^4 )) Ω = (1/e^4 )−e^3 +(9/2)+6i+(4/5)(4−3i)(e^(3i) −(1/e^4 ))](https://www.tinkutara.com/question/Q135650.png)
$$\Omega\:=\:\int_{\mathrm{C}} \left({e}^{{z}} −\overset{\_} {{z}}\right){dz} \\ $$$$\mathrm{Let}\:\mathrm{O}\left({z}=\mathrm{0}\right),\:\mathrm{A}\left(\mathrm{3}{i}\right),\:\mathrm{B}\left(−\mathrm{4}\right) \\ $$$$\Omega\:=\:\int_{\mathrm{OB}} \left({e}^{{z}} −\overset{\_} {{z}}\right){dz}+\int_{\mathrm{BA}} \left({e}^{{z}} −\overset{\_} {{z}}\right){dz}+\int_{\mathrm{AO}} \left({e}^{{z}} −\overset{\_} {{z}}\right){dz} \\ $$$$\Omega\:=\:\int_{\mathrm{OB}} \left({e}^{{z}} −\overset{\_} {{z}}\right){dz}+\int_{\mathrm{BA}} \left({e}^{{z}} −\overset{\_} {{z}}\right){dz}+\int_{\mathrm{AO}} \left({e}^{{z}} −\overset{\_} {{z}}\right){dz} \\ $$$$\mathrm{OB}\::\:{z}\:=\:{x},\:{x}\:\mathrm{goes}\:\mathrm{to}\:\mathrm{0}\:\mathrm{to}\:−\mathrm{4} \\ $$$$\mathrm{BA}\::\:{z}\:=\:{x}+{iy}\:=\:{x}\:+{i}\left(\frac{\mathrm{3}}{\mathrm{4}}{x}+\mathrm{3}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:{z}\:=\:\left(\mathrm{1}+\frac{\mathrm{3}{i}}{\mathrm{4}}\right){x}+\mathrm{3}{i} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:{x}\:\mathrm{goes}\:\mathrm{to}\:−\mathrm{4}\:\mathrm{to}\:\mathrm{0} \\ $$$$\mathrm{AO}\::\:{z}\:=\:{y},\:{y}\:\mathrm{goes}\:\mathrm{to}\:\mathrm{3}\:\mathrm{to}\:\mathrm{0} \\ $$$$\Omega\:=\:\int_{\mathrm{0}} ^{−\mathrm{4}} \left({e}^{{x}} −{x}\right){dx}+\int_{−\mathrm{4}} ^{\mathrm{0}} \left({e}^{\mathrm{3}{i}} {e}^{\left(\mathrm{1}+\frac{\mathrm{3}{i}}{\mathrm{4}}\right){x}} −\left(\mathrm{1}−\frac{\mathrm{3}{i}}{\mathrm{4}}\right){x}+\mathrm{3}{i}\right){dx} \\ $$$$+\int_{\mathrm{3}} ^{\mathrm{0}} \left({e}^{{y}} −{y}\right){dy} \\ $$$$\Omega\:=\:\left[{e}^{{x}} −\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\right]_{\mathrm{0}} ^{−\mathrm{4}} +\left[\frac{{e}^{\mathrm{3}{i}} }{\mathrm{1}+\frac{\mathrm{3}{i}}{\mathrm{4}}}{e}^{\left(\mathrm{1}+\frac{\mathrm{3}{i}}{\mathrm{4}}\right){x}} −\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}−\frac{\mathrm{3}{i}}{\mathrm{4}}\right){x}^{\mathrm{2}} +\mathrm{3}{ix}\right]_{−\mathrm{4}} ^{\mathrm{0}} \\ $$$$+\left[{e}^{{y}} −\frac{{y}^{\mathrm{2}} }{\mathrm{2}}\right]_{\mathrm{3}} ^{\mathrm{0}} \\ $$$$\Omega\:=\:\left(\frac{\mathrm{1}}{{e}^{\mathrm{4}} }−\mathrm{9}\right)+\left(\frac{{e}^{\mathrm{3}{i}} }{\mathrm{1}+\frac{\mathrm{3}{i}}{\mathrm{4}}}\left(\mathrm{1}−{e}^{−\mathrm{4}−\mathrm{3}{i}} \right)+\mathrm{8}\left(\mathrm{1}−\frac{\mathrm{3}{i}}{\mathrm{4}}\right)+\mathrm{12}{i}\right) \\ $$$$+\left(\frac{\mathrm{11}}{\mathrm{2}}−{e}^{\mathrm{3}} \right) \\ $$$$\Omega\:=\:\frac{\mathrm{1}}{{e}^{\mathrm{4}} }−{e}^{\mathrm{3}} +\frac{\mathrm{9}}{\mathrm{2}}+\mathrm{6}{i}+\frac{\mathrm{4}}{\mathrm{5}}\left(\mathrm{4}−\mathrm{3}{i}\right)\left({e}^{\mathrm{3}{i}} −\frac{\mathrm{1}}{{e}^{\mathrm{4}} }\right) \\ $$$$\Omega\:=\:\frac{\mathrm{1}}{{e}^{\mathrm{4}} }−{e}^{\mathrm{3}} +\frac{\mathrm{9}}{\mathrm{2}}+\mathrm{6}{i}+\frac{\mathrm{4}}{\mathrm{5}}\left(\mathrm{4}−\mathrm{3}{i}\right)\left({e}^{\mathrm{3}{i}} −\frac{\mathrm{1}}{{e}^{\mathrm{4}} }\right) \\ $$