The integral of the function (z/(z+9)^2)dz along the contour γ(t) = 2i + 4e^it, where 0 ≤ t ≤ 2π, evaluates to zero due to Cauchy's theorem. The function is holomorphic within the defined circle, as its singularity at -9 lies outside this contour. Therefore, the application of Cauchy's integral formula confirms that the integral yields a result of zero.
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Yes, that seems to be the case.Woolyabyss said:Homework Statement
Use Cauchy’s integral formula to evaluate the integral along γ(t) of (z/(z+9)^2)dz
where γ(t) = 2i + 4e^it , 0 ≤ t ≤ 2π.
Homework Equations
Cauchy's integral formula
The Attempt at a Solution
I was just wondering is the integral not just zero by Cauchy's theorem since (z/(z+9)^2) is holomorphic inside the circle defined by γ(t) ( the singularity at -9 is outside the circle ).