Prove that $\frac{1}{f'(a)} = \frac{1}{2\pi i} \int_C \frac{1}{f(z)-f(a)} dz$

Question

Let $f$ be analytic on an open set $U$, let $a \in U$, and $f'(a)\neq 0$.

Show that \begin{align*} \frac{1}{f'(a)} = \frac{1}{2\pi i} \int_C \frac{1}{f(z)-f(a)} dz \end{align*} where $C$ is a small circle centered at a.

I thought that this is very similar with Cauchy's integral formula. But, I can't apply the Cauchy's integral formula. How to apply that?

Any help is appreciated..

Thank you!

Answer 1

The correct formula is $$ \begin{align*} \frac{1}{f'(a)} = \frac{1}{2\pi i} \int_C \frac{1}{f(z)-f(a)} dz \, , \end{align*} $$ which holds if the interior of $C$ contains no other roots of $f(z) - f(a)$ except $z=a$.

It follows from the Residue theorem: For $z \to a$ we have $$ f(z) = f(a) + f'(a)(z-a) + O(z-a)^2 \\ \Longrightarrow \frac{1}{f(z)-f(a)} = \frac{1}{f'(a)(z-a)} + O(1) \\ \Longrightarrow \frac{1}{2\pi i} \int_C \frac{1}{f(z)-f(a)} dz = \operatorname{Res}(\frac{1}{f(z)-f(a)}, a) = \frac{1}{f'(a)} $$

Answer 2

Well, I think this equation is not generally true.

Consider $f(z)=z^2$ and take $a=8$, then:

$$R.H.S.=\frac{1}{2\pi i}\oint_C \frac{1}{z^2-16}dz=0$$ since $\frac{1}{z^2-16}$ is holomorphic in the neighborhood of $8$.

While $L.H.S.=\frac{1}{16}\ne R.H.S.$.

Answer 3

It can be viewed as Cauchy's Integral Formula applied to the inverse of $f$.

Since $f'(a)\ne0$, the Inverse Function Theorem says that there is an analytic $g$, defined in a neighborhood of $f(a)$, so that $$ f(g(z))=z\tag1 $$ Taking the derivative of $(1)$ gives $$ f'(g(z))g'(z)=1\tag2 $$ Equation $(1)$ $$ \begin{align} \frac1{2\pi i}\int_{|z|=r}\frac1{f(z)-f(a)}\mathrm{d}z &=\frac1{2\pi i}\int_{|g(z)|=r}\frac{g'(z)}{z-f(a)}\,\mathrm{d}z\tag3\\ &=g'(f(a))\tag4\\[3pt] &=\frac1{f'(a)}\tag5 \end{align} $$ Explanation:
$(3)$: substitute $z\mapsto g(z)$
$(4)$: Cauchy's Integral Formula
$(5)$: apply $(2)$ at $z=f(a)$

Answer 4

The equation holds iff $a$ is the only simple root of $f(z)=0$. $$ \begin{aligned} \frac{1}{2 \pi i}\int_C \frac{1}{f(z)-f(a)} d z = & \frac{1}{2 \pi i} 2 \pi i \operatorname{Res}\left[\frac{1}{f(z)-f(a)}, a\right] \\ = & \lim _{z \rightarrow a} \frac{z-a}{f(z)-f(a)}\\=& \frac 1{\lim _{z \rightarrow a} \frac{f(z)-f(a)}{z-a}}\\=&\frac{1}{f^{\prime}(a)} \end{aligned} $$