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[SOLVED] Analytic on U

dwsmith

Well-known member
Feb 1, 2012
1,673
Let $f$ be analytic on an open set $U$, let $z_0\in U$, and let $f'(z_0)\neq 0$. Show that
$$
\frac{2\pi i}{f'(z_0)}=\int_C\frac{1}{f(z)-f(z_0)}dz,
$$
where $C$ is some circle centered at $z_0$.

If I re-write the above expression, it is saying the winding number is 1, correct? I am not sure with what to do next or if that is the correct observation.
 

dwsmith

Well-known member
Feb 1, 2012
1,673
Let $f$ be analytic on an open set $U$, let $z_0\in U$, and let $f'(z_0)\neq 0$. Show that
$$
\frac{2\pi i}{f'(z_0)}=\int_C\frac{1}{f(z)-f(z_0)}dz,
$$
where $C$ is some circle centered at $z_0$.

If I re-write the above expression, it is saying the winding number is 1, correct? I am not sure with what to do next or if that is the correct observation.
Cauchy's Formula for derivatives

$$
f'(z_0) = \frac{1!}{2\pi i}\int_C\frac{f(z)}{(z-z_0)^2}dz
$$

Would this help?

$f(z)-f(z_0) = a_1(z-z_0)+a_2(z-z_0)^2+\cdots$ with $a_1=f'(z_0)\neq 0$
 
Last edited:

dwsmith

Well-known member
Feb 1, 2012
1,673
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
$$
\frac{1}{f'(z_0)} = \frac{z - z_0}{f(z) - f(z_0)} = \frac{1}{c_1 + c_2(z - z_0) + \cdots}.
$$
Thus there is a Taylor Series expansion, so $\frac{z - z_0}{f(z) - f(z_0)}$ is analytic on a disc at $z_0$.
From Cauchy's Theorem, we have
$$
\frac{1}{2\pi i}\int_C\frac{dz}{f(z) - f(z_0)} = \frac{1}{f'(z_0)}\Leftrightarrow
\int_C\frac{dz}{f(z) - f(z_0)} = \frac{2\pi i}{f'(z_0)}.
$$

Is this correct?