[itex]f[/itex] is analytic on an open set [itex]U[/itex], [itex]z_0\in U[/itex], and [itex]f'(z_0)\neq 0[/itex]. Show that(adsbygoogle = window.adsbygoogle || []).push({});

[tex]\frac{2\pi i}{f'(z_0)}=\int_C\frac{1}{f(z)-f(z_0)}dz[/tex]

where $C$ is some circle center at $z_0$.

S0 ,[itex]f(z)-f(z_0) = a_1(z-z_0)+a_2(z-z_0)^2+\cdots[/itex] with [itex]a_1=f'(z_0)\neq 0[/itex]. But why can [itex]f(z)-f(z_0)[/itex] be expanded this way?

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# Cauchy Integral Formula application

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