Combination Theorem for Sequences/Complex/Product Rule/Proof 1

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$\ds \lim_{n \mathop \to \infty} \paren {z_n w_n} = c d$


Because $\sequence {z_n}$ converges, it is bounded by Convergent Sequence is Bounded.

Suppose $\cmod {z_n} \le K$ for $n = 1, 2, 3, \ldots$.


\(\ds \cmod {z_n w_n - c d}\) \(=\) \(\ds \cmod {z_n w_n - z_n d + z_n d - c d}\)
\(\ds \) \(\le\) \(\ds \cmod {z_n w_n - z_n d} + \cmod {z_n d - c d}\) Triangle Inequality for Complex Numbers
\(\ds \) \(=\) \(\ds \cmod {z_n} \cmod {w_n - d} + m \cdot \size {z_n - c}\) Complex Modulus of Product of Complex Numbers
\(\ds \) \(\le\) \(\ds K \cdot \cmod {w_n - d} + \cmod d \cdot \cmod {z_n - c}\)
\(\ds \) \(=:\) \(\ds \phi_n\)

But $z_n \to c$ as $n \to \infty$.

So $\cmod {z_n - c} \to 0$ as $n \to \infty$ from Convergent Sequence Minus Limit.

Similarly $\cmod {w_n - d} \to 0$ as $n \to \infty$.

From the Combined Sum Rule for Real Sequences:

$\displaystyle \lim_{n \mathop \to \infty} \paren {\lambda z_n + \mu w_n} = \lambda c + \mu d$, $\phi_n \to 0$ as $n \to \infty$

The result follows by the Squeeze Theorem for Sequences of Complex Numbers (which applies as well to real as to complex sequences).