# Cauchy Integral Formula application

1. Feb 19, 2012

### fauboca

$f$ is analytic on an open set $U$, $z_0\in U$, 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 some circle center at $z_0$.

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

2. Feb 19, 2012

### Dick

You are given that f(z) is analytic. Doesn't that mean it can be locally expanded in a power series?

3. Feb 19, 2012

### fauboca

Yes but I don't understand how the expansion of $f(z)-f(z_0)$ is that expansion.

4. Feb 19, 2012

### Dick

f(z) is analytic at z=z0 if f(z)=a0+a1*(z-z0)+a2*(z-z0)^2+... in a neighborhood of z0? What's f(z0)?

5. Feb 19, 2012

### fauboca

The accumulation point.

6. Feb 19, 2012

### Dick

No, f(z0)=a0.

7. Feb 27, 2012

### fauboca

The Taylor Series expansion of $f(z) = \sum_{n = 0}^{\infty}c_n(z - z_0)^n$ $= c_0 + c_1(z - z_0) + c_2(z - z_0)^2 +\cdots$, and $f(z_0) = c_0$.
So,
$$f(z) - f(z_0) = c_1(z - z_0) + c_2(z - z_0)^2 +\cdots.$$
By factoring, we obtain $f(z) - f(z_0) = c_1(z - z_0)\left[1 + \frac{c_2}{c_1}(z - z_0)^2 +\cdots\right]$.
Then
$$\frac{z - z_0}{f(z) - f(z_0)} = \frac{1}{c_1 + c_2(z - z_0) + \cdots}.$$
So $g(z) = \frac{z - z_0}{f(z) - f(z_0)}$ is analytic on a disc (what is the justification for this part-analytic?) at $z_0$ (How does this disc relate to C?) and where the value of $z_0$ is redefined as $1/c_1$.
From Cauchy's Theorem, we have
$$\frac{1}{2\pi i}\int_C\frac{g(z)}{z - z_0} = g(z_0) = \frac{1}{2\pi i}\int_C\frac{dz}{f(z) - f(z_0)} = \frac{1}{f'(z_0)}$$
hence
$$\int_C\frac{dz}{f(z) - f(z_0)} = \frac{2\pi i}{f'(z_0)}.$$

8. Feb 27, 2012

### Dick

Your g(z) has a removable singularity at z=z0. And the problem just says "some circle centered at z0". Take the circle small enough so that g(z) is analytic in the circle.

9. Feb 27, 2012

Ok thanks.