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

	Answered by mathmax by abdo last updated on 11/Jun/21
	$Φ=∫_C ((cos(πz^2 )+sin(πz^2 ))/((z+1)(z+2)))dz let Ψ(z)=((cos(πz^2 )+sin(πz^2 ))/((z+1)(z+2))) the[poles[of Ψ are −1 and −2 ⇒ ∫_C Ψ(z)dz =2iπ {Res(Ψ,−1)+Res(Ψ ,−2)} Res(Ψ,−1)=((cos(π)+sin(π))/1)=−1 Res(Ψ,−2) =((cos(4π)+sin(4π))/(−1))=−1 ⇒ ∫_C Ψ(z)dz =2iπ(−1−1) =−4iπ ⇒Φ=−4iπ$
	$Φ = \int_{C} \frac{\cos (π z^{2}) + \sin (π z^{2})}{(z + 1) (z + 2)} dz let Ψ (z) = \frac{\cos (π z^{2}) + \sin (π z^{2})}{(z + 1) (z + 2)}$ $the [poles [of Ψ are - 1 and - 2 \Rightarrow$ $\int_{C} Ψ (z) dz = 2 i π {Res (Ψ, - 1) + Res (Ψ, - 2)}$ $Res (Ψ, - 1) = \frac{\cos (π) + \sin (π)}{1} = - 1$ $Res (Ψ, - 2) = \frac{\cos (4 π) + \sin (4 π)}{- 1} = - 1 \Rightarrow$ $\int_{C} Ψ (z) dz = 2 i π (- 1 - 1) = - 4 i π \Rightarrow Φ = - 4 i π$
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