(1)∫ ((cosz)/((z+1)^2 ))dz , ∣z∣=1 (2) ∫ ((cosz)/((z−1)^2 ))dz ,∣z∣=1


Question and Answers Forum

All Questions Topic List
None Questions
Previous in All Question Next in All Question
Previous in None Next in None
Question Number 147213 by Mrsof last updated on 18/Jul/21

$(1) \int \frac{c o s z}{{(z + 1)}^{2}} d z, ∣ z ∣= 1$ $(2) \int \frac{c o s z}{{(z - 1)}^{2}} d z, ∣ z ∣= 1$
	Answered by mathmax by abdo last updated on 19/Jul/21
	$I=∫_C ((cosz)/((z+1)^2 )) with C→∣z∣=1 f(z)=((cosz)/((z+1)^2 )) residus ⇒∫_C f(z)dz=2iπ Res(f,−1) Res(f,−1)=lim_(z→−1) (1/((2−1)!)){(z+1)^2 f(z)}^((1)) =lim_(z→−1) {cosz}^((1)) =lim_(z→−1) (−sinz) =sin(1) ⇒I=sin(1)$
	$I = \int_{C} \frac{cosz}{{(z + 1)}^{2}} with C \to∣ z ∣= 1$ $f (z) = \frac{cosz}{{(z + 1)}^{2}} residus \Rightarrow \int_{C} f (z) dz = 2 i π Res (f, - 1)$ $Res (f, - 1) = \lim_{z \to - 1} \frac{1}{(2 - 1)!} {{(z + 1)}^{2} f (z)}^{(1)}$ $= \lim_{z \to - 1} {cosz}^{(1)} = \lim_{z \to - 1} (- sinz) = \sin (1) \Rightarrow I = \sin (1)$
	Answered by mathmax by abdo last updated on 19/Jul/21

	$2) \int_{C} \frac{cosz}{{(z - 1)}^{2}} dz =_{z = - w} - \int_{C} \frac{\cos (w)}{{(w + 1)}^{2}} dw = - \sin (1)$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum