continuity in terms of u and v
Here is the most useful fact about complex continuity: it reduces to something you already know from multivariable calculus. Write f = u + i v with u(x, y) and v(x, y) the real and imaginary parts. Then f is continuous at the point z_0 = x_0 + i y_0 exactly when both real functions u and v are continuous at (x_0, y_0) in the ordinary two-variable sense. Complex continuity is componentwise continuity.
Why this works comes straight from the modulus. The size of the change in f is |f(z) - f(z_0)|, and this is the length of the vector whose components are the change in u and the change in v. That length is small precisely when both component-changes are small — that is the content of the inequalities |u - u_0| <= |f - f_0| and |v - v_0| <= |f - f_0|, alongside |f - f_0| <= |u - u_0| + |v - v_0|. So f stays near f(z_0) if and only if u and v each stay near their values.
This is liberating in practice: to check continuity of a complex function you never need new machinery — just confirm that two ordinary real functions of (x, y) are continuous, using all the familiar tools (continuity of polynomials, the algebra of limits, the squeeze theorem). For example f(z) = z^2 has u = x^2 - y^2 and v = 2 x y, both visibly continuous, so z^2 is continuous everywhere. The same translation underlies the deeper Cauchy-Riemann story, where u and v are studied as a coupled pair.
For f(z) = e^z, separating gives u = e^x cos(y) and v = e^x sin(y). Each is a product and composition of continuous real functions, so both are continuous everywhere; therefore e^z is continuous on the whole plane.
Checking continuity of f reduces to checking continuity of the two real functions u and v.
This equivalence is only about continuity, not differentiability. Both u and v can be perfectly smooth real functions while f fails to be complex-differentiable — that further requires the Cauchy-Riemann equations linking u and v, not just each one being nice on its own.