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I'm reviewing HW problems and had a question.

The problem statement is the following:

Let $f$ be analytic on an open set $U$, let $z_0 \in U$ and $f'(z_0) \neq 0$. Show that $$\frac{2 \pi i}{f'(z_0)} = \int_C \frac{1}{f(z)-f(z_0)} dz$$ where C is a small circle centered at $z_0$.

First I found an online solution but I don't understand one step. The solution writes $f(z)-f(z_0) = a_1(z-z_0)(1+\frac{a_2}{a_1}(z-z_0)+ \dots)$ and I agree.

Next it says since $\frac{z-z_0}{f(z)-f(z_0)}$ is analytic so by Cauchy's fomula: $$\frac{1}{2 \pi i}\int_C \frac{1}{f(z)-f(z_0)} dz = \frac{1}{a_1} = \frac{1}{f'(z_0)}$$

I don't understand how to get $$\frac{1}{2 \pi i}\int_C \frac{1}{f(z)-f(z_0)} dz = \frac{1}{a_1}$$

Also, I wonder since Cauchy's formula writes $$\frac{1}{2 \pi i} \int_C\frac{f(z)}{z-z_0}dz = f(z_0)$$ Why can't I let $f$ in Cauchy's formula be $\frac{z-z_0}{f(z)-f(z_0)}$ so $f(z_0)$ in Cauchy's formula by L'Hospital Rule equals $\frac{1}{f'(z_0)}$?

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The Cauchy Integral formula says $$ g(z_0) = \frac{1}{2\pi i}\int_C \frac{g(z)}{z-z_0}dz $$ where $g(z)$ is analytic inside $C$. In your case $$ \frac{1}{f(z)-f(z_0)}=\frac{1}{(z-z_0)(f'(z_0)+\frac{1}{2}f''(z_0)(z-z_0)+\cdots)} $$ Now consider the function $$ g(z) = \frac{1}{f'(z_0)+\frac{1}{2}f''(z_0)(z-z_0)+\cdots}:=[h(z)]^{-1} $$ The point is to show that $g(z)$ has no poles inside $C$. Here is what smallness of the circle comes into the picture. First of all since $f'(z_0)\neq 0$ we do not have any poles for $g(z)$ on $z_0$. Suppose the sequence of zeroes of $h(z)$ is denoted by $\zeta_n$, i.e. $h(\zeta_n)=0$. If this sequence converges to $z_0$ then by continuity of $h(z)$ (or $f(z)$) we will have $h(z_0)=0$ which is a contradiction. So we have a open neighborhood around $z_0$ not containing any zeroes of $h(z)$. Take $C$ small enough such that it is contained in this open neighborhood. Then $g(z)$ is analytic inside $C$. Therefore $$ \frac{1}{2\pi i}\int_C \frac{g(z)}{z-z_0}dz = g(z_0)=\frac{1}{f'(z_0)} $$

For your second question choosing $F(z) = \frac{z -z_0}{f(z)-f(z_0)}$ and using Cauchy integral formula can work, but you basically want to show $$ F(z_0)=\int_C \frac{F(z)}{z-z_0}dz = \lim_{z\to z_0}F(z) $$ By l'Hopital. So first you need to show $F(z)$ is well-defined and analytic insider $C$. Proving these statements is equivalent to previous argument.

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  • $\begingroup$ Nice explanation! Seems I did not clearly realize what are the things require checking. Thank you. $\endgroup$ – MonkeyKing Nov 16 '15 at 18:05
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I don't see what the confusion is. The solution you've proposed is effectively the same thing as the one you have read. You use the Taylor series to prove that

$$ g(z) = \frac{z-z_0}{f(z)-f(z_0)}\equiv \frac{1}{a_1(1+\frac{a_1}{a_2}(z-z_0)+\ldots)} $$ is equivalent to an analytic function. Then you apply Cauchy to $g(z)$ at $z=z_0$ to obtain

$$ \frac{1}{2\pi i}\oint_C\frac{1}{f(z)-f(z_0)}dz = \frac{1}{2\pi i}\oint_C\frac{g(z)}{z-z_0}dz = g(z_0) = \frac{1}{a_1} = \frac{1}{f^\prime(z_0)} $$ A more powerful way to approach your problem is the theory of Residues, which you may or may not be familiar with.

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