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