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joypav

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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 $\omega$?

"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 $\omega$?

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