Complex functions, limits, and continuity

Setting up the plane. Write $z=x+iy$ and let a function act as $f(z)=f(x+iy)=u(x,y)+i\,v(x,y)$, so one complex rule packages two real-valued functions $u,v:\mathbb{R}^2\to\mathbb{R}$. Distance in $\mathbb{C}$ is the ordinary Euclidean distance $|z_1-z_2|=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$, so $\mathbb{C}$ behaves exactly like $\mathbb{R}^2$ for the purposes of limits.

Limit. The formal definition mirrors the real $\varepsilon$-$\delta$ definition, only the absolute value now measures planar distance:

where: $\varepsilon$ is the target closeness of $f(z)$ to $L$, $\delta$ is the resulting closeness of $z$ to $z_0$ that guarantees it, and $|z-z_0|$ is the modulus of the difference, i.e. planar distance. The decisive difference from $\mathbb{R}$ is that the disc $0<|z-z_0|<\delta$ surrounds $z_0$ in every direction; the definition demands the same $L$ along every path of approach, not merely along the axes.

Continuity. $f$ is continuous at $z_0$ when three things all hold: $f(z_0)$ is defined, $\lim_{z\to z_0}f(z)$ exists, and

Component reduction. The practical shortcut, proved below, is that continuity of $f$ reduces to continuity of two ordinary real multivariable functions $u(x,y)$ and $v(x,y)$ at the point $(x_0,y_0)$ — a fact you can check with tools already familiar from the several-variables unit.

Algebra of limits. Sums, differences, products, and quotients (nonzero denominator limit) of complex functions obey the same limit laws as in $\mathbb{R}$. A direct consequence: every polynomial in $z$ is continuous on all of $\mathbb{C}$, and every rational function is continuous wherever its denominator does not vanish.

Problem. Extend the Pass-1 examples: using the component-reduction theorem, verify directly (via $u,v$) that $f(z)=z^2+3z$ is continuous at $z_0=1+i$ with $f(1+i)=3+5i$, and confirm that $g(z)=\dfrac{z+1}{z^2+1}$ is discontinuous only at $z=\pm i$.

Step 1. Split $f(z)=z^2+3z$ into real and imaginary parts. With $z=x+iy$: $z^2=(x^2-y^2)+i(2xy)$, so $f(z)=(x^2-y^2+3x)+i(2xy+3y)$, giving $u(x,y)=x^2-y^2+3x$ and $v(x,y)=2xy+3y$.

Step 2. Both $u$ and $v$ are polynomials in the real variables $x,y$, and polynomials of several real variables are continuous everywhere on $\mathbb{R}^2$ (sums and products of the continuous coordinate functions $x,y$). By the theorem just derived, $f$ is therefore continuous at every point of $\mathbb{C}$, in particular at $z_0=1+i$ (so $x_0=1,\,y_0=1$).

Step 3. Evaluate at $(x_0,y_0)=(1,1)$: $u(1,1)=1-1+3=3$ and $v(1,1)=2+3=5$, so $f(1+i)=3+5i$ — matching the direct complex computation $f(1+i)=(1+i)^2+3(1+i)=(1+2i-1)+3+3i=3+5i$ from Pass-1, now confirmed through the $u,v$ route.

Step 4. For $g(z)=\dfrac{z+1}{z^2+1}$, the numerator and denominator are each polynomials in $z$, hence continuous everywhere by the same reasoning as Step 2 applied to their own real and imaginary parts. A quotient of continuous functions is continuous except where the denominator is zero, and $z^2+1=0$ gives $z=\pm i$. So $g$ is continuous on $\mathbb{C}\setminus\{i,-i\}$ and undefined (hence discontinuous) exactly at these two points.

Answer: $f(z)=z^2+3z$ is continuous everywhere, with $f(1+i)=3+5i$ confirmed by both the direct and the $u,v$ methods; $g(z)=(z+1)/(z^2+1)$ is discontinuous only at $z=i$ and $z=-i$.

• Checking only the two axis directions ($y=0$ or $x=0$) and declaring a limit "exists" — the definition demands agreement along every path, as the real-axis-versus-imaginary-axis test on $\bar z/z$ shows (limits $1$ and $-1$, so no limit exists).
• Treating $|z|=\sqrt{x^2+y^2}$ as if it behaves like a differentiable complex function just because it is continuous everywhere; continuity and (complex) differentiability are very different strengths, and $|z|$ is nowhere complex-differentiable — a later topic, not this one.
• Forgetting to write $f(z)=u(x,y)+iv(x,y)$ explicitly before invoking the component-reduction theorem; skipping the split leads to sign or variable-mixing errors when computing $u$ and $v$.
• Assuming a rational function is continuous everywhere without checking the denominator; it fails exactly at the finitely many roots of the denominator, as with $g(z)$ at $z=\pm i$.

Ask the tutor: Want more two-path counterexamples like $\bar z/z$, or a walkthrough splitting a trickier function like $e^z$ into its $u,v$ parts, ask the AI tutor to generate practice problems on complex limits and continuity.