Continuity of complex function

A complex function $f$ is continuous at $z_0$ if

${\lim }_{z\to z_0}f(z)=f(z_0).$

That is, the limit exists, $f$ is defined at $z_0$, and the two agree. $f$ is continuous on a set if it’s continuous at every point of the set.

Reduction to real-variable continuity

Writing $f(z)=u(x,y)+iv(x,y)$, $f$ is continuous at $z_0=x_0+i{y}_0$ iff both $u$ and $v$ are continuous as functions of two real variables at $(x_0,y_0)$. So complex continuity reduces to a pair of standard multivariable continuity statements.

What’s continuous

• Constants and the identity $f(z)=z$.
• $\bar{z}$, $|z|$, $\mathrm{Re}(z)$, $\mathrm{Im}(z)$.
• Polynomials (sums of $a_n{z}^n$): continuous everywhere.
• Rational functions $p(z)/q(z)$: continuous everywhere $q(z)\ne 0$.
• $e^z$, $\sin z$, $\cos z$: continuous everywhere.
• Sums, products, compositions, and quotients (with nonzero denominator) of continuous functions.

A continuous-but-not-differentiable function

$f(z)=\bar{z}$ is continuous everywhere on $\mathbb{C}$: both $u(x,y)=x$ and $v(x,y)=-y$ are continuous as functions of $(x,y)$.

Yet $\bar{z}$ is complex differentiable nowhere. Continuity is much weaker than differentiability in the complex setting.

Contrast the real-variable case, where continuous-but-nowhere-differentiable functions (like the Weierstrass function) are pathological and rare. In the complex setting that behaviour is the norm: any smooth function involving $\bar{z}$, $\mathrm{Re}(z)$, $\mathrm{Im}(z)$, or $|z|$ in a real way will be continuous but not differentiable.

Differentiability implies continuity

The converse fails badly (continuity does not imply differentiability), but complex differentiability does imply continuity:

${\lim }_{z\to z_0}(f(z)-f(z_0))={\lim }_{z\to z_0}\frac{f(z)-f(z_0)}{z-z_0}(z-z_0)=f^{\prime }(z_0)\cdot 0=0.$

So any analytic function is continuous on its domain of analyticity.

Where it sits

Continuity is the weakest regularity worth having. The Complex derivative needs more, and Cauchy’s theorem needs full analyticity. But you can’t define a contour integral of $f$ without continuity along the contour, so this is the floor.