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Question Number 135635 by mohammad17 last updated on 14/Mar/21

Commented by mohammad17 last updated on 14/Mar/21

$$helpmesir$$

Answered by Olaf last updated on 14/Mar/21

$$\mathrm{\Omega }={\int }_{\mathrm{C}}(e^z-\stackrel{\mathrm{\_}}{z})dz$$$$\mathrm{Let}\mathrm{O}(z=0),\mathrm{A}\left(3i\right),\mathrm{B}(-4)$$$$\mathrm{\Omega }={\int }_{\mathrm{OB}}(e^z-\stackrel{\mathrm{\_}}{z})dz+{\int }_{\mathrm{BA}}(e^z-\stackrel{\mathrm{\_}}{z})dz+{\int }_{\mathrm{AO}}(e^z-\stackrel{\mathrm{\_}}{z})dz$$$$\mathrm{\Omega }={\int }_{\mathrm{OB}}(e^z-\stackrel{\mathrm{\_}}{z})dz+{\int }_{\mathrm{BA}}(e^z-\stackrel{\mathrm{\_}}{z})dz+{\int }_{\mathrm{AO}}(e^z-\stackrel{\mathrm{\_}}{z})dz$$$$\mathrm{OB}:z=x,x\mathrm{goes}\mathrm{to}0\mathrm{to}-4$$$$\mathrm{BA}:z=x+iy=x+i(\frac{3}{4}x+3)$$$$z=(1+\frac{3i}{4})x+3i$$$$x\mathrm{goes}\mathrm{to}-4\mathrm{to}0$$$$\mathrm{AO}:z=y,y\mathrm{goes}\mathrm{to}3\mathrm{to}0$$$$\mathrm{\Omega }={\int }_0^{-4}(e^x-x)dx+{\int }_{-4}^0(e^{3i}{e}^{(1+\frac{3i}{4})x}-(1-\frac{3i}{4})x+3i)dx$$$$+{\int }_3^0(e^y-y)dy$$$$\mathrm{\Omega }={[e^x-\frac{x^2}{2}]}_0^{-4}+{[\frac{e^{3i}}{1+\frac{3i}{4}}{e}^{(1+\frac{3i}{4})x}-\frac{1}{2}(1-\frac{3i}{4})x^2+3ix]}_{-4}^0$$$$+{[e^y-\frac{y^2}{2}]}_3^0$$$$\mathrm{\Omega }=(\frac{1}{e^4}-9)+(\frac{e^{3i}}{1+\frac{3i}{4}}(1-e^{-4-3i})+8(1-\frac{3i}{4})+12i)$$$$+(\frac{11}{2}-e^3)$$$$\mathrm{\Omega }=\frac{1}{e^4}-e^3+\frac{9}{2}+6i+\frac{4}{5}(4-3i)(e^{3i}-\frac{1}{e^4})$$$$\mathrm{\Omega }=\frac{1}{e^4}-e^3+\frac{9}{2}+6i+\frac{4}{5}(4-3i)(e^{3i}-\frac{1}{e^4})$$

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