Question-143238

Question Number 143238 by BHOOPENDRA last updated on 11/Jun/21

Answered by mathmax by abdo last updated on 11/Jun/21

Φ=∫_C ((cos(πz^2 )+sin(πz^2 ))/((z+1)(z+2)))dz let Ψ(z)=((cos(πz^2 )+sin(πz^2 ))/((z+1)(z+2))) the[poles[of Ψ are −1 and −2 ⇒ ∫_C Ψ(z)dz =2iπ {Res(Ψ,−1)+Res(Ψ ,−2)} Res(Ψ,−1)=((cos(π)+sin(π))/1)=−1 Res(Ψ,−2) =((cos(4π)+sin(4π))/(−1))=−1 ⇒ ∫_C Ψ(z)dz =2iπ(−1−1) =−4iπ ⇒Φ=−4iπ

$$\Phi=\int_{\mathrm{C}} \frac{\mathrm{cos}\left(\pi\mathrm{z}^{\mathrm{2}} \right)+\mathrm{sin}\left(\pi\mathrm{z}^{\mathrm{2}} \right)}{\left(\mathrm{z}+\mathrm{1}\right)\left(\mathrm{z}+\mathrm{2}\right)}\mathrm{dz}\:\:\mathrm{let}\:\Psi\left(\mathrm{z}\right)=\frac{\mathrm{cos}\left(\pi\mathrm{z}^{\mathrm{2}} \right)+\mathrm{sin}\left(\pi\mathrm{z}^{\mathrm{2}} \right)}{\left(\mathrm{z}+\mathrm{1}\right)\left(\mathrm{z}+\mathrm{2}\right)} \\ $$$$\mathrm{the}\left[\mathrm{poles}\left[\mathrm{of}\:\Psi\:\mathrm{are}\:−\mathrm{1}\:\mathrm{and}\:−\mathrm{2}\:\Rightarrow\right.\right. \\ $$$$\int_{\mathrm{C}} \Psi\left(\mathrm{z}\right)\mathrm{dz}\:=\mathrm{2i}\pi\:\left\{\mathrm{Res}\left(\Psi,−\mathrm{1}\right)+\mathrm{Res}\left(\Psi\:,−\mathrm{2}\right)\right\} \\ $$$$\mathrm{Res}\left(\Psi,−\mathrm{1}\right)=\frac{\mathrm{cos}\left(\pi\right)+\mathrm{sin}\left(\pi\right)}{\mathrm{1}}=−\mathrm{1} \\ $$$$\mathrm{Res}\left(\Psi,−\mathrm{2}\right)\:=\frac{\mathrm{cos}\left(\mathrm{4}\pi\right)+\mathrm{sin}\left(\mathrm{4}\pi\right)}{−\mathrm{1}}=−\mathrm{1}\:\Rightarrow \\ $$$$\int_{\mathrm{C}} \Psi\left(\mathrm{z}\right)\mathrm{dz}\:=\mathrm{2i}\pi\left(−\mathrm{1}−\mathrm{1}\right)\:=−\mathrm{4i}\pi\:\Rightarrow\Phi=−\mathrm{4i}\pi \\ $$

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