Complex continuity

Three things must hold at \(z_0\): \(f(z_0)\) exists, the limit exists, and they agree. Failing any of them produces a discontinuity; if the limit exists but is not \(f(z_0)\) it is called removable, because redefining the value eliminates it. Thus \((z^2-4)/(z-2)\) with \(f(2)=c\) is continuous at \(2\) exactly when \(c=4\).

Polynomials, \(\operatorname{Re}\), \(\operatorname{Im}\), \(|z|\) and \(\bar z\) are continuous on all of \(\mathbb{C}\). Sums, products, quotients (where the denominator does not vanish) and compositions of continuous functions are continuous, so \(1/(z^2-4)\) is continuous on \(\mathbb{C}\setminus\{\pm 2\}\). A function \(f=u+iv\) is continuous at \(z_0\) if and only if \(u\) and \(v\) are at \((x_0,y_0)\).

Continuity is much weaker than complex differentiability: \(\bar z\) is continuous everywhere and, as will be seen, differentiable nowhere. It does inherit the theorems of real functions: on a closed bounded set, such as the disk \(|z|\le 1\), a continuous function is bounded and \(|f|\) attains its maximum; and if \(f(z_0)\neq 0\), it stays nonzero near \(z_0\).

A frequent mistake is to prove continuity by checking only some paths, when the definition asks for a single bound valid for all of them. The next step is the complex exponential, the first genuinely new continuous function.