Proof of Winding Number 1 for Analytic $f$

In summary, the conversation discusses the relationship between a function $f$ that is analytic on an open set $U$, a point $z_0\in U$, and the derivative of $f$ at $z_0$. It is shown that if $f'(z_0)\neq 0$, then $\frac{2\pi i}{f'(z_0)}$ can be represented as the integral of $\frac{1}{f(z)-f(z_0)}$ along some circle $C$ centered at $z_0$. The conversation also explores the use of Cauchy's Formula for derivatives and the Taylor Series expansion of $f$ to arrive at this result.
  • #1
Dustinsfl
2,281
5
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.
 
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  • #2
dwsmith said:
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:
  • #3
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?
 

FAQ: Proof of Winding Number 1 for Analytic $f$

What is the winding number of an analytic function?

The winding number of an analytic function is a measure of how many times the function winds around a point in the complex plane. It is a way of quantifying the behavior of a function near a point.

Why is a winding number of 1 important for analytic functions?

A winding number of 1 for an analytic function means that the function is injective, or one-to-one, in a neighborhood of the point in question. This is important because it guarantees that the function has a well-defined inverse in that neighborhood.

How is the winding number of a function related to its derivative?

The winding number of a function is related to its derivative by the argument principle. The argument principle states that the winding number of a function around a point is equal to the change in the argument of the function around that point divided by 2π. This can be used to calculate the winding number by integrating the derivative of the function around the point.

Can the winding number of a function be negative or zero?

Yes, the winding number of a function can be negative or zero. A negative winding number indicates that the function winds around the point in the opposite direction, while a winding number of zero indicates that the function does not wind around the point at all.

What are some real-world applications of the winding number?

The winding number has applications in many areas of mathematics and physics, including complex analysis, topology, and fluid dynamics. It is also used in computer graphics and image processing to analyze and manipulate images. In addition, the winding number has practical applications in engineering, such as in the design of electrical circuits and the analysis of control systems.

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