(z_n) Converges IFF (Re(z_n)) and (Im(z_n)) Converges

# A Sequence of Complex Numbers Converges IFF The Real Part Sequence and Imaginary Part Sequence Converges

Proposition 1: Let $(z_n)_{n=1}^{\infty} = (x_n + i y_n)_{n=1}^{\infty}$ be a sequence of complex numbers. Then $(z_n)_{n=1}^{\infty}$ converges to $Z = X + Yi$ if and only if $(x_n)_{n=1}^{\infty}$ converges to $X$ and $(y_n)_{n=1}^{\infty}$ converges to $Y$. |

**Proof:**$\Rightarrow$ Suppose that $(x_n + iy_n)_{n=1}^{\infty}$ converges to $X + Yi$. Then for all $\epsilon > 0$ there exists an $N \in \mathbb{N}$ such that if $n \geq N$ then:

\begin{align} \quad \mid z_n - Z \mid = \mid x_n + iy_n - (X + Yi) \mid = \mid x_n - X + i(y_n - Y) \mid = \sqrt{(x_n - X)^2 + (y_n - Y)^2} < \epsilon \end{align}

- For this $N$ we have that if $n \geq N$ then:

\begin{align} \quad \mid x_n - X \mid = \sqrt{(x_n - X)^2} < \sqrt{(x_n - X) + (y_n - Y)^2} < \epsilon \end{align}

(3)
\begin{align} \quad \mid y_n - Y \mid = \sqrt{(y_n - Y)^2} < \sqrt{(x_n - X) + (y_n - Y)^2} < \epsilon \end{align}

- Therefore $(x_n)_{n=1}^{\infty}$ converges to $X$ and $(y_n)_{n=1}^{\infty}$ converges to $Y$.

- $\Leftarrow$ Conversely, suppose that $(x_n)_{n=1}^{\infty}$ converges to $X$ and $(y_n)_{n=1}^{\infty}$ converges to $Y$]. Let $\epsilon > 0$ be given. Then there exists $N_1, N_2 \in \mathbb{N}$ such that if $n \geq N_1$ we have that $\mid x_n - X \mid < \frac{\epsilon}{2}$ and if $n \geq N_2$ we have that $\mid y_n - Y \mid < \frac{\epsilon}{2}$. Let $N = \max \{ N_1, N_2 \}$. Then if $n \geq N$ we have that:

\begin{align} \quad \quad \mid z_n - Z \mid = \mid (x_n + y_ni) - (X + Yi) \mid = \mid (x_n - X) + i(y_n - Y) \mid \leq \mid x_n - X \mid + \mid y_n - Y \mid < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon \end{align}

- Therefore $(z_n)_{n=1}^{\infty}$ converges to $Z$. $\blacksquare$