Complex Variables - Zeros of Analytic Functions

[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 $\omega$?

 

[Euge]

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

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

 

[joypav]

Got it... thank you

 

[joypav]

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?

 

[Euge]

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 $f^{-1}(w)$.