# Complex Variables - Zeros of Analytic Functions

• MHB
• joypav
In summary, the conversation discusses the extension of a formula to prove a generalized theorem involving analytic functions. It also touches on the method of proof using factorization and integration, and the concept of biholomorphisms. The ultimate goal is to show that $f^{-1}(w)$ can be expressed as an integral involving $f$ and its derivative, and the conversation ends with a clarification on how the right side of the integral can be equivalent to $f^{-1}(w)$.
joypav
Studying for my complex analysis final. I think this should be a simple question but wanted some clarification.

"Extend the formula

$$\frac{1}{2i\pi} \int_\omega \frac{h'(z)}{h(z)}\, dz = \sum_{j=1}^N n_j - \sum_{k=1}^M m_k$$

to prove the following.

Let $g$ be analytic on a domain containing $\omega$ and its inside. Then

$$\frac{1}{2i\pi} \int_\omega \frac{h'(z)}{h(z)}g(z)dz = \sum_{i=1}^N g(z_i) - \sum_{j=1}^M g(w_j)$$

where $z_1,...,z_N$ are the zeros of h and $w_1,...,w_M$ are the poles of h inside $\omega$, each listed according to its multiplicity."

Does this just utilize the theorem that states that

$\displaystyle \frac{1}{2i\pi} \int_\omega \frac{h'(z)}{h(z)}\, dz =$ number of zeros of h inside $\omega$ - number of poles of h inside $\omega$?

Last edited by a moderator:
Hi joypav,

A method of proof used to prove the argument principle applies to the generalized theorem. Namely, factorize

$$h(z) = \frac{(z - z_1)\cdots (z - z_N)}{(z - w_1)\cdots (z - w_M)} f(z)$$

where $f$ is analytic and zero-free on an open neighborhood of $\omega$.

$$\frac{h'(z)}{h(z)}g(z) = \sum_{i = 1}^N \frac{g(z)}{z - z_i} - \sum_{j = 1}^M \frac{g(z)}{z - w_j} + \frac{f'(z)}{f(z)}g(z)$$

Integrate both sides over $\omega$, divide by $2 i\pi$, and use Cauchy's integral formula and Cauchy's theorem to establish

$$\frac{1}{2i\pi} \int_\omega \frac{h'(z)}{h(z)} = \sum_{i = 1}^N g(z_i) - \sum_{j = 1}^M g(w_j)$$

Got it... thank you

Following this same question... I want to show that
$$\displaystyle f^{-1}(w)=\frac{1}{2\pi i} \int_\omega \frac{zf'(z)}{f(z)-w}\, dz$$
where f is analytic and one to one on a domain D.

What I have:

$\omega$ is a piecewise smooth closed curve in D whose inside lies in D. Say $\Omega=f(D)$.
(the final goal is to show that $f^{-1}$ is analytic on $\Omega$)

Let w be a point of $\Omega$.
Let $g(z)=z$ and $h(z)=f(z)-w$.

We know from the previous question,
$$\frac{1}{2\pi i} \int_\omega \frac{h'(z)}{h(z)}g(z)dz = \sum_{i=1}^N g(z_i) - \sum_{j=1}^M g(w_j)$$

By making substitutions we get,
$$\frac{1}{2\pi i} \int_\omega \frac{f'(z)}{f(z)-w}zdz = \sum_{i=1}^N z_i - \sum_{j=1}^M w_j$$Now it seems like I am almost done. Is this correct so far?
I'm not sure how why the right side is equivalent to $f^{-1}(w)$, or if I am even on the right track.
We know,
$$h(z)=f(z)+w$$
$$f(z)=h(z)-w$$
$$f^{-1}(z)=h^{-1}(z-w)$$
$$f^{-1}(w)=h^{-1}(0)$$
And h has zeros ${z_1,...,z_N}$
I thought that may be how I can show their equivalence?

The inverse function $f^{-1}$ need not be analytic. In fact, analytic functions with analytic inverses are called biholomorphisms (some call them analytic isomorphisms).

The injectivity of $f$ implies $f(z) - w$ has a unique zero, say $z_0$. Since $f$ is analytic, $f(z) - w$ has no poles. Furthermore, $f'(z)$ equals the derivative of $f(z) - w$. So by the generalized argument principle, your integral equals $z_0$, which is $f^{-1}(w)$.

## 1. What are complex variables and how are they related to zeros of analytic functions?

Complex variables are numbers that have both a real and an imaginary component. They are related to zeros of analytic functions because these functions are defined and studied using complex variables. The zeros of an analytic function are the values of its input variable (usually represented as z) that make the function output a value of 0. In other words, they are the solutions to the equation f(z) = 0.

## 2. How do analytic functions differ from other types of functions?

Analytic functions are different from other types of functions because they are defined and studied using complex variables. They are also known as holomorphic functions and have a number of unique properties, such as being infinitely differentiable and having a Taylor series expansion. These properties make them useful for solving complex problems in mathematics, physics, and engineering.

## 3. Can a complex variable have more than one zero for an analytic function?

Yes, a complex variable can have multiple zeros for an analytic function. In fact, an analytic function can have an infinite number of zeros in the complex plane. This is because the zeros of an analytic function are determined by the behavior of the function near its singularities, which can be infinitely close together in the complex plane.

## 4. How can the zeros of an analytic function be found?

The zeros of an analytic function can be found by using various methods, such as graphing the function in the complex plane, using numerical methods like the Newton-Raphson method or the bisection method, or using algebraic techniques like the quadratic formula. Additionally, the behavior of the function near its singularities can provide valuable information about its zeros.

## 5. What is the significance of the zeros of an analytic function?

The zeros of an analytic function have a number of important applications in mathematics and other fields. For example, they can be used to solve equations and systems of equations, to find the roots of polynomials, and to study the behavior of complex systems. In physics, the zeros of analytic functions are related to the poles of the associated physical systems, which can provide insights into the behavior of these systems.

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