Sir cu module

[Beniamin Bogosel]

Demonstrati ca daca \( z \) si \( w \) sunt numere complexe diferite aflate in interiorul discului unitate, atunci \( |\frac{z^n-w^n}{z-w}-n\cdot w^{n-1}|\to 0 \).

[Theodor Munteanu]

Avem \( |z| < 1,\ |w| < 1,\ z \ne w \);

\( \frac{{z^n - w^n }}{{z - w}} = z^{n - 1} + z^{n - 2} w + ...zw^{n - 2} + w^{n - 1} \);

\( |z^{n - 1} + z^{n - 2} w + ...zw^{n - 2} + w^{n - 1} - nw^{n - 1} | \le \\
|z|^{n - 1} + |z|^{n - 2} |w| + |z|^{n - 3} |w|^2 + ... + |z||w|^{n - 2} + |w|^{n - 1} + n|w|^{n - 1} ; \\
\)

Trecand la limita
\( \lim \limits_{n \to \infty } (|z|^{n - 1} + |z|^{n - 2} |w| + |z|^{n - 3} |w|^2 + ... + |z||w|^{n - 2} + |w|^{n - 1}+ n|w|^{n - 1} ) = \lim \limits_{n \to \infty } |z|^{n - 1} + \lim\limits_{n \to \infty } |z|^{n - 2} |w| + ... \)

Dar \( \lim \limits_{n \to \infty } |z|^{n - k} |w|^{k - 1} = 0 \) deoarece \( |w|^{{\rm k - 1}} < 1 \) iar \( \lim \limits_{n \to \infty } |z|^{n - k} = 0 \).

Deci limita initiala e 0.