Prove: z_n→ a if and only if Re\z_n\→Re\a\ and Im\z_n\→Im\a\

Prove: $z_n\to a$ if and only if $\text{Re}\{z_n\}\to\text{Re}\{a\}$ and $\text{Im}\{z_n\}\to\text{Im}\{a\}$.

Direction $(\Rightarrow)$: $z_n\to a$ implies convergence of both parts

Since $z_n-a=(x_n-\alpha)+i(y_n-\beta)$:

$$ |x_n-\alpha|\le\sqrt{(x_n-\alpha)^2+(y_n-\beta)^2}=|z_n-a|, \qquad |y_n-\beta|\le|z_n-a| $$

Given $\varepsilon>0$, choose $N$ with $|z_n-a|<\varepsilon$ for $n>N$; then both $|x_n-\alpha|<\varepsilon$ and $|y_n-\beta|<\varepsilon$ for $n>N$, so $x_n\to\alpha,\ y_n\to\beta$.

Direction $(\Leftarrow)$: convergence of both parts implies $z_n\to a$

Using the triangle inequality $\sqrt{p^2+q^2}\le|p|+|q|$:

$$ |z_n-a|\le|x_n-\alpha|+|y_n-\beta| $$

Given $\varepsilon>0$, choose $N_1$ with $|x_n-\alpha|<\varepsilon/2$ for $n>N_1$, and $N_2$ with $|y_n-\beta|<\varepsilon/2$ for $n>N_2$. Then for $n>N=\max(N_1,N_2)$:

$$ |z_n-a|\le|x_n-\alpha|+|y_n-\beta|<\frac\varepsilon2+\frac\varepsilon2=\varepsilon $$

so $z_n\to a$.

Why this works · equivalence of norms

The argument rests on $\max(|x|,|y|)\le\sqrt{x^2+y^2}\le|x|+|y|\le2\max(|x|,|y|)$ — the modulus and the "taxicab" combination of real/imaginary distances are equivalent measures of closeness in $\mathbb C\cong\mathbb R^2$, so convergence in one sense is exactly convergence in the other.