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If f(z) and (z-a) are holomorphic functions in D(0, r) and if $\gamma(t) = re^{it}, 0≤ ≤2pi$.

Then, using the Cauchy integral formula we have $\dfrac{1}{a-b} \int_{\gamma} \dfrac{f(z)}{z-a} = f(a)2\pi i / (a-b)$

I don't understand that equality

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    How did $(a-b)$ get in there? As for the integral being a multiple of $f(a)$, notice that $f(a)$ is the 0-th term in the Taylor series of $f(z)$ at $z=a$. See Cartan's "Elementary Theory of Analytic Functions". – 2017-01-26

2 Answers 2

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Cauchy's integral formula says $\int_\gamma \frac{f(z)}{z-a}\, dz =2\pi i f(a)$. Divide both sides by $a-b$ you should have your expression.

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Cauchy's integral formula says that if $f$ is holomorphic in $B_r(0)$, then $$ f(a)=\frac{1}{2 \pi i}\int_{\partial B_r(0)}\frac{f(z)}{z-a}dz$$ for all $a \in B_r(0).$

$\gamma(t)=re^{it}$, $0 \leq t \leq 2 \pi$ is just a parametrization of $\partial B_r(0)$. So $$ \frac{2 \pi if(a)}{a-b}=\frac{1}{a-b}\int_{\gamma}\frac{f(z)}{z-a}dz$$ works only if $a \in B_r(0)$ and $b\neq a.$