Complex functions, limits, and differentiability in 2D

The language we need before we can talk about limits

Before the phrase "the limit of $f(z)$ as $z$ approaches $z_0$" can mean anything precise, you need to be able to say what kind of region $z$ is approaching $z_0$ inside. A small vocabulary, defined exam-style.

Open disk (or neighbourhood) of $z_0$ with radius $\rho > 0$:

The set of all points strictly closer to $z_0$ than $\rho$. The boundary circle $|z - z_0| = \rho$ is not included. Sometimes called the $\rho$-neighbourhood of $z_0$.

Open set. A set $S \subseteq \mathbb{C}$ is open if, around every point $z_0 \in S$, you can find a neighbourhood — some open disk — lying entirely inside $S$. The interior of a circle is open; a disk that includes its boundary is not.

Closed set. A set is closed if its complement is open — equivalently, if it contains all of its boundary points. The disk $|z| \le 1$ is closed; the disk $|z| < 1$ is open.

Boundary. A point $z_0$ lies on the boundary of $S$ if every neighbourhood of $z_0$ contains both points in $S$ and points outside $S$. The circle $|z| = 1$ is the boundary of both $|z| < 1$ and $|z| \le 1$.

Domain. A set is a domain if it is open and connected — any two of its points can be joined by a path lying inside the set. A region is a domain together with some, all, or none of its boundary points. "Domain" is the workhorse word: limits, continuity, and analyticity are all stated relative to domains in the chapters ahead.

One worked sketch

The set $S = \{\, z : 1 < |z - i| < 2 \,\}$ is an open annulus around $z = i$, with inner radius $1$ and outer radius $2$. It is open — around any point inside, you can fit a small disk. It is connected — you can walk from any point to any other along an arc. So it is a domain.

Its boundary is the union of the two circles $|z - i| = 1$ and $|z - i| = 2$. Add those circles in and you get a closed annulus — a region that is no longer open.

Limits in two dimensions, and the canonical failure pattern

The informal definition, which is the one you will actually use: $\lim_{z \to z_0} f(z) = L$ means as $z$ approaches $z_0$ from any direction, the value $f(z)$ approaches $L$. That last phrase — from any direction — is the only thing new compared with real-line limits, and it is exactly the thing that makes complex limits interesting.

How limits actually get computed: split into real and imaginary parts

Write $f(z) = u(x, y) + i\,v(x, y)$ and $z_0 = x_0 + iy_0$. Then:

Read that twice: a complex limit is just two real limits in two variables, one for the real part and one for the imaginary part. Both must exist for the complex limit to exist, and the limit is then their combination. This is the rule that turns limit computations into routine work. The standard algebra of limits carries straight over:

Worked example — a limit that exists

Compute $\lim_{z \to i} (z^2 + z + 1)$. Since $z^2 + z + 1$ is a polynomial — and polynomials are continuous everywhere — you can just substitute:

The complex limit is $i$. Splitting into parts as a check: the real part of $f$ is $u(x,y) = x^2 - y^2 + x + 1$, and at $(x,y) = (0,1)$, $u = 0 - 1 + 0 + 1 = 0$. The imaginary part is $v(x,y) = 2xy + y$, and at $(0,1)$, $v = 0 + 1 = 1$. So $\lim u + i\lim v = 0 + i\cdot 1 = i$. Same answer.

Worked example — a limit that fails (the central pattern of the chapter)

Compute $\lim_{z \to 0} \dfrac{z}{\bar z}$. You cannot substitute — at $z = 0$, both $z$ and $\bar z$ are zero, so the expression is the indeterminate $0/0$. We have to actually take the limit. Try two paths.

Path 1 — along the real axis ($y = 0$, so $z = x$ and $\bar z = x$):

Path 2 — along the imaginary axis ($x = 0$, so $z = iy$ and $\bar z = -iy$):

Two different paths gave two different answers — $1$ and $-1$. A complex limit must give the same answer along every path; since these two disagree, the limit does not exist.

Continuity

A function $f$ is continuous at $z_0$ if $\lim_{z \to z_0} f(z) = f(z_0)$ — and, of course, $f(z_0)$ has to actually be defined. A function is continuous on a domain if it is continuous at every point of the domain. The split-into-parts trick still applies: $f = u + iv$ is continuous at $z_0 = x_0 + iy_0$ exactly when both $u(x,y)$ and $v(x,y)$ are continuous at $(x_0, y_0)$.

Polynomials are continuous everywhere. Rational functions are continuous wherever their denominator is non-zero. $\bar z$ is continuous everywhere — conjugation preserves distances, $|\bar z - \bar{z_0}| = |z - z_0|$. The exponential $e^z$ and the trigonometric functions $\sin z$, $\cos z$ are continuous everywhere too; you will meet them properly in the harmonic-conjugates chapter.

Why differentiability is more than the limit existing

A function $f$ is differentiable at $z_0$ if the following limit exists:

Here $h$ is a complex number going to zero — meaning $|h| \to 0$ — and $h$ can approach zero from any direction in the complex plane. That clause is the whole content of this chapter. In real calculus, $h \to 0$ means from the left or from the right — two directions, and you only insist the two one-sided limits agree. In complex calculus, $h$ can go to zero along the positive real axis, along the imaginary axis, along the line $y = x$, along a spiral, along a parabola. For $f'(z_0)$ to exist, all of these must give the same complex number. So "differentiable" in the complex sense is dramatically stronger than in the real sense — it is a constraint on the function from infinitely many directions at once.

Worked example — $\bar z$ is differentiable nowhere

Take $f(z) = \bar z$ and try to compute $f'(z_0)$ at any point $z_0$:

Notice that $z_0$ has disappeared — whatever happens here happens at every point in the same way. The question is now: does $\bar h / h$ have a limit as $h \to 0$?

Approach 1 — along the real axis ($h = a$ with $a \in \mathbb{R}$, $a \to 0$):

Approach 2 — along the imaginary axis ($h = ib$ with $b \in \mathbb{R}$, $b \to 0$):

Two paths, two different answers. So the limit defining $f'(z_0)$ does not exist — and the same argument runs at every $z_0$. $\bar z$ is differentiable nowhere in the complex plane.

Worked example — $z^2$ is differentiable, with $f'(z) = 2z$

To contrast, here is a function that works. Take $f(z) = z^2$. Then

As $h \to 0$ this approaches $2 z_0$ — regardless of the direction $h$ approaches from, because the leftover $h$ term goes to zero from every direction at the same rate. So $f'(z_0) = 2 z_0$, exactly as in real calculus, and $z^2$ is differentiable everywhere in the complex plane. Why does it succeed where $\bar z$ fails? Because in the algebra the direction of $h$ never leaked through to the final answer. With $\bar z$, the conjugation pinned the direction into the formula and the limit picked it up; with $z^2$, the direction got squeezed out. The Cauchy-Riemann chapter explains which functions have this happy property — and gives you a two-line test you can run on any $f$.

The playground — a direction-of-approach visualizer

A live tool to make the every-direction demand visible. Choose a function and a point, rotate the direction of approach with the $\theta$ slider, and watch the difference quotient either hold steady (differentiable) or swing around (not differentiable). The tool does not differentiate symbolically — it numerically evaluates the difference quotient with a tiny $|h| = 10^{-6}$, which is precisely what the chapter is teaching.

Work through this sequence — the pattern across the four is the real lesson:

1. Start with $\bar z$ at $z_0 = 1 + i$. Slide $\theta$ from $0°$ to $360°$ and watch the difference quotient move around the unit circle. The pattern repeats every $180°$ — that is the $e^{-2i\theta}$ invariant from the "failure is total" card above.
2. Switch to $z^2$. Slide $\theta$ again. The difference quotient stays fixed at $2 z_0 = 2 + 2i$, exactly what the worked example predicted.
3. Try $|z|^2$ at $z_0 = 1 + i$ and note the failure. Then set $z_0 = 0 + 0i$ and slide $\theta$ again — now it works, and the value is $0$. That is the "differentiable at one point only" phenomenon you will prove yourself in Reveal 2 below.
4. Try $1/z$ at $z_0 = 1 + 0i$. The answer stays put at $-1$. A sneak preview: $1/z$ is analytic everywhere except at $z = 0$, and the Cauchy-Riemann chapter confirms it.

Test yourself before moving on

Do not peek. Commit to an answer before you reveal anything. Committing — even to a wrong answer — is what builds the memory. Recognising a correct answer once you see it only feels like learning.

How this concept appears on exams

The material in this chapter shows up on engineering exams in three recurring shapes. None is hard once you have seen the template.

Shape 1 — "Show that the limit does not exist."

A common 4-mark question. Always solved by the path-failure pattern:

1. State the function and the point $z_0$.
2. Pick path 1 (real axis, $y = 0$). Substitute, simplify, state the limit along path 1.
3. Pick path 2 (imaginary axis, $x = 0$, or the line $y = x$ if the axes do not separate). State the limit along path 2.
4. Observe: the two limits differ.
5. Conclude: the limit does not exist.

The marks come from the structure, not the choice of paths. Many students lose marks for skipping step 4 (the explicit observation) or step 5 (the explicit conclusion). Write them down.

Shape 2 — "From first principles, show whether $f$ is differentiable, and if so find $f'$."

A 5- or 6-mark question. Template:

1. Write the difference quotient $\bigl(f(z_0 + h) - f(z_0)\bigr)/h$ and expand the numerator.
2. Simplify, separating any $\bar h$ from $h$.
3. If a $\bar h / h$ survives, pick two directions and show different limits. Conclude: not differentiable.
4. If the algebra leaves only terms like $2 z_0$, $\bar{z_0}$, or $h$ (which goes to zero), conclude the limit is the constant term and state $f'(z_0)$.

The technique is mechanical once you have done it twice. The exam habit to drill is always separating into $h$ and $\bar h$ early — students who go straight to $h = a + ib$ and never group the $\bar h$ terms build themselves algebra cliffs.

Shape 3 — "Is $f$ continuous? Is $f$ differentiable?"

A common conceptual trap. The student sees a function — often $\bar z$, $|z|$, $|z|^2$, or $\operatorname{Re}(z)$ — notices it is "clearly nice", meaning continuous, and writes "yes, differentiable" without thinking. The standard examiner trap is to hand you a function that is continuous everywhere but differentiable nowhere, or differentiable at one point only. Reliable examples in each category:

• Continuous everywhere, differentiable nowhere: $\bar z$, $\operatorname{Re}(z)$, $\operatorname{Im}(z)$, $|z|$, $2x + iy$.
• Differentiable only at $z = 0$: $|z|^2$, $\bar z^{\,2}$ (worth checking by the same method as Reveal 2).
• Differentiable everywhere (and analytic): polynomials in $z$, $e^z$, $\sin z$, $\cos z$, rational functions away from their poles.

The habit that saves marks: before answering "differentiable", compute the difference quotient and look for $\bar h$. If $\bar h$ shows up, you have a path-dependence problem. Ten seconds on that check recovers three marks.

What you can do now

1. Write a complex function $f(z) = u(x,y) + i\,v(x,y)$ and identify its real and imaginary parts.
2. Define an open disk, open set, closed set, domain, and region, and classify simple sets like $|z| < 1$, $1 < |z - i| < 2$, or $\operatorname{Im}(z) > 0$.
3. State the informal definition of $\lim_{z \to z_0} f(z) = L$, and the $\varepsilon$-$\delta$ definition for completeness.
4. Use the split rule $\lim f = \lim u + i\lim v$ to compute existing complex limits, and use the two-path failure pattern to show a limit does not exist.
5. Define continuity at a point and on a domain, and recognise that polynomials and rational functions are continuous on their domains.
6. Define the complex derivative as a limit of the difference quotient $\bigl(f(z_0 + h) - f(z_0)\bigr)/h$, and say clearly why its existence requires the same value along every direction of approach.
7. Compute $f'(z_0)$ from first principles for $z^2$ and other polynomials.
8. Show that $\bar z$ is differentiable nowhere by exhibiting two paths with different limits, and that $|z|^2$ is differentiable only at $z = 0$.
9. Recognise the continuous-everywhere-but-differentiable-nowhere trap and answer it without losing marks.

Same material, another voice

If a different explanation would help, this one is worth your time — free, from MIT OpenCourseWare:

• Watch: Herbert Gross, Calculus Revisited, Lecture 2 — Functions of a Complex Variable — a friendly chalkboard walk through complex functions, limits, and the complex derivative.

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