Combination Theorem for Sequences/Complex/Sum Rule/Proof 2

Theorem
Let $\sequence {z_n}$ and $\sequence {w_n}$ be sequences in $\C$.

Let $\sequence {z_n}$ and $\sequence {w_n}$ be convergent to the following limits:


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

Then:

Proof
Let $\epsilon > 0$ be given.

Then $\dfrac \epsilon 2 > 0$.

We are given that:
 * $\ds \lim_{n \mathop \to \infty} z_n = c$
 * $\ds \lim_{n \mathop \to \infty} w_n = d$

Let:
 * $z_n = x_n + i y_n$
 * $w_n = r_n + i s_n$


 * $c = a + i b$
 * $d = l + i m$

where each of $x_n, y_n, r_n, s_n, a, b, l, m \in \R$ are real.

By definition:

$\sequence {z_n}$ converges to the limit $c = a + i b$ both:


 * $\forall \epsilon \in \R_{>0}: \exists N \in \R: n > N \implies \size {x_n - a} < \epsilon \text { and } \size {y_n - b} < \epsilon$

$\sequence {w_n}$ converges to the limit $d = l + i m$ both:


 * $\forall \epsilon \in \R_{>0}: \exists N \in \R: n > N \implies \size {r_n - l} < \epsilon \text { and } \size {s_n - m} < \epsilon$

where $\size x$ denotes the absolute value of $x \in \R$.

Then: