Hints for proving that $\lim_{z \to z_0} g(z) = a + bi$ implies that the real and imaginary parts converge to $a$ and $b$?

Question

Let $S \subseteq \mathbb{C}$ and $g: S \to \mathbb{C}$

Premises: $$\lim_{z \to z_0} g(z) = a + bi \tag {1}$$

$$\forall (z \in \text{dom g }) g(z)= u(z) + i v(z) \tag {2}$$

I would like to show that these premises imply that $\lim_{z \to z_0} u(z) = a$ and $\lim_{z \to z_0} v(z) = b$.

I wrote out and manipulated the limit definition of premise $(1)$:

$\forall (\epsilon > 0) \exists (\delta > 0) \forall (z \in S): 0 < |z - z_0| < \delta \implies |(u(z)-a) + i(v(z)-b)| = 0$.

From this, I can see that the converse would be very easy to prove using the triangle inequality, but I'm really not sure what to do to prove this direction.

I think that if I tried I can prove that if either $u$ or $v$ converges to $a$ or $b$, then the other one must converge too. But I don't think this is very useful. I would appreciate any ideas.

I would also like to ask if premise $(2)$ is true for any function $g: S \to \mathbb{C}.$

Answer 1

Hint: Use the very useful fact that if $z = a + ib$ (cartesian form) then $$ \max\{|a|,|b|\} \leq |z| $$