Combination Theorem for Sequences/Complex/Multiple Rule
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Theorem
Let $\sequence {z_n}$ be a sequence in $\C$.
Let $\sequence {z_n}$ be convergent to the following limit:
- $\ds \lim_{n \mathop \to \infty} z_n = c$
Let $\lambda \in \C$.
Then:
- $\ds \lim_{n \mathop \to \infty} \paren {\lambda z_n} = \lambda c$
Proof
Let $\epsilon > 0$.
We need to find $N$ such that:
- $\forall n > N: \cmod {\lambda z_n - \lambda c} < \epsilon$
If $\lambda = 0$ the result is trivial.
So, assume $\lambda \ne 0$.
Then $\cmod \lambda > 0$ from the definition of the modulus of $\lambda$.
Hence $\dfrac \epsilon {\cmod \lambda} > 0$.
We have that $z_n \to c$ as $n \to \infty$.
Thus it follows that:
- $\exists N: \forall n > N: \cmod {z_n - c} < \dfrac \epsilon {\cmod \lambda}$
That is:
- $\forall n > N: \cmod \lambda \cmod {z_n - c} < \epsilon$
But we have:
\(\ds \cmod \lambda \cmod {z_n - c}\) | \(=\) | \(\ds \cmod {\lambda \paren {z_n - c} }\) | Complex Modulus of Product of Complex Numbers | |||||||||||
\(\ds \) | \(=\) | \(\ds \cmod {\lambda x_n - \lambda l}\) |
Hence:
- $\ds \lim_{n \mathop \to \infty} \paren {\lambda x_n} = \lambda c$
$\blacksquare$