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Question Number 129688 by Engr_Jidda last updated on 17/Jan/21

complex analysis  ∮_C ((ϱ^(2z) +sinz^2 )/((z−2)(z−3)))dz   C:∣Z∣=5

$${complex}\:{analysis} \\ $$$$\oint_{{C}} \frac{\varrho^{\mathrm{2}{z}} +{sinz}^{\mathrm{2}} }{\left({z}−\mathrm{2}\right)\left({z}−\mathrm{3}\right)}{dz}\:\:\:{C}:\mid{Z}\mid=\mathrm{5} \\ $$

Answered by mathmax by abdo last updated on 17/Jan/21

ϕ(z)=((e^(2z)  +sin(z^2 ))/((z−2)(z−3))) the poles of ϕ are 2 and 3  ∫_(∣z∣=5)   ϕ(z)dz=2iπ{ Res(ϕ,2) +Res(ϕ,3)}  =2iπ{((e^4 +sin(4))/(−1)) +((e^6  +sin9)/1)} =2iπ{e^6 −e^4  +sin9−sin4}

$$\varphi\left(\mathrm{z}\right)=\frac{\mathrm{e}^{\mathrm{2z}} \:+\mathrm{sin}\left(\mathrm{z}^{\mathrm{2}} \right)}{\left(\mathrm{z}−\mathrm{2}\right)\left(\mathrm{z}−\mathrm{3}\right)}\:\mathrm{the}\:\mathrm{poles}\:\mathrm{of}\:\varphi\:\mathrm{are}\:\mathrm{2}\:\mathrm{and}\:\mathrm{3} \\ $$$$\int_{\mid\mathrm{z}\mid=\mathrm{5}} \:\:\varphi\left(\mathrm{z}\right)\mathrm{dz}=\mathrm{2i}\pi\left\{\:\mathrm{Res}\left(\varphi,\mathrm{2}\right)\:+\mathrm{Res}\left(\varphi,\mathrm{3}\right)\right\} \\ $$$$=\mathrm{2i}\pi\left\{\frac{\mathrm{e}^{\mathrm{4}} +\mathrm{sin}\left(\mathrm{4}\right)}{−\mathrm{1}}\:+\frac{\mathrm{e}^{\mathrm{6}} \:+\mathrm{sin9}}{\mathrm{1}}\right\}\:=\mathrm{2i}\pi\left\{\mathrm{e}^{\mathrm{6}} −\mathrm{e}^{\mathrm{4}} \:+\mathrm{sin9}−\mathrm{sin4}\right\} \\ $$

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