find ∫_(∣z∣=1) ((1−cosz)/z^2 )dz


Question and Answers Forum

All Questions Topic List
Integration Questions
Previous in All Question Next in All Question
Previous in Integration Next in Integration
Question Number 130392 by mathmax by abdo last updated on 25/Jan/21

$find \int_{∣ z ∣= 1} \frac{1 - cosz}{z^{2}} dz$
	Answered by mohammad17 last updated on 25/Jan/21

	$h e l l o s i r c a n y o u h e l p m e i n Q 130416$
	Answered by mathmax by abdo last updated on 25/Jan/21
	$∫_(∣z∣=1) ((1−cosz)/z^2 )dz =2iπ Res(f,0) with f(z)=((1−cosz)/z^2 ) o is double pole ⇒Res(f,o)=lim_(z→0) (1/((2−1)!)){z^2 f(z)}^((1)) =lim_(z→0) (1−cosz)^((1)) =lim_(z→0) sinz =0 another way ∣z∣=1 ⇒z=e^(iθ) ⇒∫_(∣z∣=1) ((1−cosz)/z^2 ) =∫_0 ^(2π) ((1−cos(e^(iθ) ))/e^(2iθ) )ie^(iθ) dθ =i∫_0 ^(2π) e^(−iθ) (1−cos(e^(iθ) ))dθ =i∫_0 ^(2π) e^(−iθ) dθ−i∫_0 ^(2π) e^(−iθ) cos(e^(iθ) )dθ =0−i∫_0 ^(2π) e^(−iθ) (Σ_(n=0) ^∞ (((−1)^n e^(2inθ) )/((2n)!)))dθ =−iΣ_(n=0) ^∞ (((−1)^n )/(2n!)) ∫_0 ^(2π) e^((2n−1)iθ) dθ =−iΣ_(n=0) ^∞ (((−1)^n )/((2n)!))[(1/((2n−1)))e^((2n−1)iθ) ]_0 ^(2π) =0$
	$\int_{∣ z ∣= 1} \frac{1 - cosz}{z^{2}} dz = 2 i π Res (f, 0) with f (z) = \frac{1 - cosz}{z^{2}}$ $o is double pole \Rightarrow Res (f, o) = \lim_{z \to 0} \frac{1}{(2 - 1)!} {z^{2} f (z)}^{(1)}$ $= \lim_{z \to 0} {(1 - cosz)}^{(1)} = \lim_{z \to 0} sinz = 0$ $another way ∣ z ∣= 1 \Rightarrow z = e^{i θ} \Rightarrow \int_{∣ z ∣= 1} \frac{1 - cosz}{z^{2}} = \int_{0}^{2 π} \frac{1 - \cos (e^{i θ})}{e^{2 i θ}} {ie}^{i θ} d θ$ $= i \int_{0}^{2 π} e^{- i θ} (1 - \cos (e^{i θ})) d θ = i \int_{0}^{2 π} e^{- i θ} d θ - i \int_{0}^{2 π} e^{- i θ} \cos (e^{i θ}) d θ$ $= 0 - i \int_{0}^{2 π} e^{- i θ} (\sum_{n = 0}^{\infty} \frac{{(- 1)}^{n} e^{2 in θ}}{(2 n)!}) d θ$ $= - i \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{2 n!} \int_{0}^{2 π} e^{(2 n - 1) i θ} d θ$ $= - i \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{(2 n)!} {[\frac{1}{(2 n - 1)} e^{(2 n - 1) i θ}]}_{0}^{2 π} = 0$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum