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V. Simple Application of Cauchy Integral Formula

I'm puzzled by a solution given to a simple Cauchy integral formula question. The question is to evaluate ∫γcot⁡z dz\int _{\gamma } \cot z \ dz for γ=eit\gamma = e^{it}.
The solution just lets f=zcot⁡zf = z\cot z and w=0, but then the integral equals 2πif(0)2\pi i f(0). The solution claims that this is equal to 2πi2\pi i, but isn't f(0)=0⋅cos⁡0sin⁡0f(0)= 0\cdot \frac{\cos 0}{\sin 0} undefined?!

Is the solution messed up or have I forgotten to consider the particular case where I am certifiably insane?
This doesn't look Cauchy at all. :frown:
Reply 2
Avatar for Zhen Lin
Zhen Lin
12
Well, cot⁡z\cot z has a simple pole at 0. So zcot⁡zz \cot z has a removable singularity at 0. So we can (indeed, must - since Cauchy's integral formula requires f to be holomorphic on the interior of the contour) take f(0)=lim⁡z→0zcot⁡z=1\displaystyle f(0) = \lim_{z \to 0} z \cot z = 1.
Reply 3
Avatar for Kolya
Kolya
OP
17
Zhen Lin
Well, cot⁡z\cot z has a simple pole at 0. So zcot⁡zz \cot z has a removable singularity at 0. So we can (indeed, must - since Cauchy's integral formula requires f to be holomorphic on the interior of the contour) take f(0)=lim⁡z→0zcot⁡z=1\displaystyle f(0) = \lim_{z \to 0} z \cot z = 1.

Orite. ty :yy:

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