# Combination Theorem for Sequences/Complex/Multiple Rule

## Theorem

Let $\sequence {z_n}$ be a sequence in $\C$.

Let $\sequence {z_n}$ be convergent to the following limit:

$\displaystyle \lim_{n \mathop \to \infty} z_n = c$

Let $\lambda \in \C$.

Then:

$\displaystyle \lim_{n \mathop \to \infty} \paren {\lambda z_n} = \lambda c$

## 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:

 $\displaystyle \cmod \lambda \cmod {z_n - c}$ $=$ $\displaystyle \cmod {\lambda \paren {z_n - c} }$ Complex Modulus of Product of Complex Numbers $\displaystyle$ $=$ $\displaystyle \cmod {\lambda x_n - \lambda l}$

Hence:

$\displaystyle \lim_{n \mathop \to \infty} \paren {\lambda x_n} = \lambda c$

$\blacksquare$