Question Number 115009 by arcana last updated on 22/Sep/20

∫_C (e^z /(1−cos z))dz ; C:∣z∣=1

Answered by Olaf last updated on 24/Sep/20

∫_C ((coz+isinz)/(1−cosz))dz ∫_C ((1+isinz+(cosz−1))/(1−cosz))dz ∫_C (dz/(1−cosz))+i∫_C ((sinz)/(1−cosz))−∫_C dz ∫_C ((2dz)/(sin^2 (z/2)))+i∫_C ((sinz)/(1−cosz))−∫_C dz [−4cot(z/2)+iln(1−cosz)−z]_C Let z = e^(iθ) , θ∈[0 ; 2π] [−4cot(e^(iθ) /2)+iln(1−cose^(iθ) )−e^(iθ) ]_0 ^(2π) = 0

Answered by Bird last updated on 24/Sep/20

i give this solution but not sure I =∫_C (e^z /(1−cosz))dz we have 1−cosz ∼(z^2 /2) ⇒(e^z /(1−cosz))∼((2e^z )/z^2 ) so o is a double pole for f(z)=(e^z /(1−cosz)) I =2iπ ×Res(f,0) Res(f,o) =lim_(z→0) (1/((2−1)!)){z^2 (e^z /(1−cosz))}^((1)) =lim_(z→0) {2z^2 e^z }^((1)) =lim_(z→0) 2{2z e^z +2z^2 e^z } =0