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Question Number 213604 by issac last updated on 10/Nov/24
show that ∫_C e^z^3 dz=0 where C is any simple closed contour. Evaluate the integral ∫_( C_1 ) f(z)dz , ∫_( C_2 ) f(z)dz where f(z)=(y−x)−3x^2 i C_3 =OA ; z(y)=x+iy=iy , (0≤y≤1) C_1 =AB ; z(x)=x+iy=x+i , (0≤x≤1) C_2 =OB ; z(x)=x+iy=x+ix , (0≤x≤1) Let′s C be the quadrant z=2e^(iθ) ,0≤θ≤(π/2) show that ∣∫_( C) ((z^ +4)/(z^3 −1)) dz∣≤((6π)/7) Let C be any simple closed contour described in the positive sense in the z plane and write g(z)=∫_( C) ((s^3 +2s)/((z−2s)^3 )) ds show that g(z)=6πiz when z is inside C show that g(z)=0 when z is outside C
$$\mathrm{show}\:\mathrm{that}\:\:\int_{\boldsymbol{\mathcal{C}}} \:{e}^{{z}^{\mathrm{3}} } \:\mathrm{d}{z}=\mathrm{0} \\ $$$$\mathrm{where}\:\boldsymbol{\mathcal{C}}\:\mathrm{is}\:\mathrm{any}\:\mathrm{simple}\:\mathrm{closed}\:\mathrm{contour}. \\ $$$$\:\:\:\: \\ $$$$\mathrm{Evaluate}\:\mathrm{the}\:\mathrm{integral} \\ $$$$\int_{\:{C}_{\mathrm{1}} } {f}\left({z}\right)\mathrm{d}{z}\:,\:\int_{\:{C}_{\mathrm{2}} } {f}\left({z}\right)\mathrm{d}{z} \\ $$$$\mathrm{where}\:{f}\left({z}\right)=\left({y}−{x}\right)−\mathrm{3}{x}^{\mathrm{2}} \boldsymbol{{i}} \\ $$$${C}_{\mathrm{3}} ={OA}\:;\:{z}\left({y}\right)={x}+\boldsymbol{{i}}{y}=\boldsymbol{{i}}{y}\:,\:\left(\mathrm{0}\leq{y}\leq\mathrm{1}\right) \\ $$$${C}_{\mathrm{1}} ={AB}\:;\:{z}\left({x}\right)={x}+\boldsymbol{{i}}{y}={x}+\boldsymbol{{i}}\:,\:\left(\mathrm{0}\leq{x}\leq\mathrm{1}\right) \\ $$$${C}_{\mathrm{2}} ={OB}\:;\:{z}\left({x}\right)={x}+\boldsymbol{{i}}{y}={x}+\boldsymbol{{i}}{x}\:,\:\left(\mathrm{0}\leq{x}\leq\mathrm{1}\right) \\ $$$$\: \\ $$$$\mathrm{Let}'\mathrm{s}\:{C}\:\mathrm{be}\:\mathrm{the}\:\mathrm{quadrant} \\ $$$${z}=\mathrm{2}{e}^{\boldsymbol{{i}}\theta} \:,\mathrm{0}\leq\theta\leq\frac{\pi}{\mathrm{2}} \\ $$$$\mathrm{show}\:\mathrm{that}\: \\ $$$$\mid\int_{\:{C}} \:\frac{{z}^{\:} +\mathrm{4}}{{z}^{\mathrm{3}} −\mathrm{1}}\:\mathrm{d}{z}\mid\leq\frac{\mathrm{6}\pi}{\mathrm{7}} \\ $$$$\: \\ $$$$\mathrm{Let}\:{C}\:\mathrm{be}\:\mathrm{any}\:\mathrm{simple}\:\mathrm{closed}\:\mathrm{contour} \\ $$$$\mathrm{described}\:\mathrm{in}\:\mathrm{the}\:\mathrm{positive}\:\mathrm{sense}\:\mathrm{in}\:\mathrm{the}\:{z}\:\mathrm{plane} \\ $$$$\mathrm{and}\:\mathrm{write} \\ $$$$\mathrm{g}\left({z}\right)=\int_{\:{C}} \:\:\frac{{s}^{\mathrm{3}} +\mathrm{2}{s}}{\left({z}−\mathrm{2}{s}\right)^{\mathrm{3}} }\:\mathrm{d}{s} \\ $$$$\mathrm{show}\:\mathrm{that}\:\mathrm{g}\left({z}\right)=\mathrm{6}\pi\boldsymbol{{i}}{z}\:\mathrm{when}\:{z}\:\mathrm{is}\:\mathrm{inside}\:{C}\: \\ $$$$\mathrm{show}\:\mathrm{that}\:\mathrm{g}\left({z}\right)=\mathrm{0}\:\mathrm{when}\:{z}\:\mathrm{is}\:\mathrm{outside}\:{C} \\ $$

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