How does the triangle inequality prove continuous?

Question

In my Complex Analysis textbook it states that a function $f$ of the complex argument $z$ is continuous iff it is continuous viewed as a function of the two real variables $x$ and $y$. Then it states that by the triangle inequality, it is immediately known that $f$ is continuous. My question is how do we immediately know that?

Answer 1

I assume what you mean is that $f$ is continuous iff $u$ and $v$ are continuous where $f=u+iv$.

Suppose $u$ and $v$ are continuous. Let $\varepsilon>0$ and choose $\delta$ such that $|u(z)-u(a)|<\varepsilon/2$ and $|v(z)-v(a)|<\varepsilon/2$ on $B(a,\delta)$, then we also have $$|f(z)-f(a)|=|u(z)+iv(z)-(u(a)+iv(a))|=|u(z)-u(a)+i(v(z)-v(a))|\leq$$$$|u(z)-u(a)|+|v(z)-v(a)|<\varepsilon/2+\varepsilon/2.$$