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I want to find $f(1)$ and $f'(3)$, where $$f(z) = \dfrac{1}{2\pi i} \int_{|\zeta| = 2} \dfrac {e^{1/({\zeta - 5})}}{\zeta - z} d \zeta$$ where the integral is taken over a positively oriented circle. I think, that $f'(3) = 0$ because of Cauchy theorem. But I have some problems with $f(1)$.

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    $\begingroup$ $f(1)=e^{1/(1-5)}$ by Residue Theorem. $\endgroup$ – Kavi Rama Murthy Dec 25 '20 at 12:28
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$$f(z)=\frac1{2\pi i}\oint_{|\zeta|=2}\frac{e^{1/(\zeta-5)}}{\zeta-z}d\zeta\implies f'(z)=\frac1{2\pi i}\oint_{|\zeta|=2}\frac d{dz}\left(\frac{e^{1/(\zeta-5)}}{\zeta-z}\right)d\zeta=$$

$$=\frac1{2\pi i}\oint_{|\zeta|=2}\frac{e^{1/(\zeta-5)}}{(\zeta-z)^2}d\zeta=\frac{2\pi i}{1!\cdot2\pi i}\frac d{dz}\left(e^{1/(z-5)}\right)=\left(-\frac1{(z-5)^2}\,e^{1/(z-5)}\right)\implies$$

$$f'(3)=-\frac1{(-2)^2}\,e^{-1/2}=-\frac1{4\sqrt e}$$

Observe that the last expression in the first line and the first expression in the second line is just the integral theorem by Cauchy

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