Limits and Continuity in the Complex Plane

In this lesson you’ll state the epsilon-delta definition of a complex limit, see why “from every direction” is the phrase that matters, learn the path test for proving a limit does not exist, and check continuity by splitting into real and imaginary parts.

$\displaystyle\lim_{z\to z_0}f(z) = L$ means: for every $\varepsilon>0$ there is $\delta>0$ such that

$$0 < |z-z_0| < \delta \quad\Longrightarrow\quad |f(z)-L| < \varepsilon$$

Word for word this is the real-analysis definition. Only the meaning of the absolute value has changed, and that change is the whole story.

In $\mathbb{R}$, the set $|x - x_0| < \delta$ is an interval, and $x$ can approach $x_0$ from two sides. In $\mathbb{C}$, the set $|z-z_0|<\delta$ is a disc, and $z$ can approach $z_0$ along infinitely many paths: any ray, any spiral, any wiggling curve.

The limit must be the same along every one of them. That is a much stronger requirement, and it makes limits easier to disprove and harder to satisfy.

The practical consequence is a clean way to show a limit fails to exist:

If two paths approaching $z_0$ give different limiting values, then $\lim_{z\to z_0}f(z)$ does not exist.

Rays are usually the easiest paths to try, because on a ray you can substitute $z = te^{i\theta}$ and let $t\to0^+$ with $\theta$ fixed. If the answer still mentions $\theta$, you are finished.

The reference example is $f(z) = \bar z/z$. On the ray $\arg z = \theta$,

$$\frac{\bar z}{z} = \frac{te^{-i\theta}}{te^{i\theta}} = e^{-2i\theta}$$

which is constant along each ray and different on different rays. Along the real axis the value is 1; along the imaginary axis it is $-1$. The limit at 0 does not exist, even though $|f(z)| = 1$ everywhere, so the function is bounded and never wild.

This is the same phenomenon as the multivariable limits you saw in Calculus 3, where $\frac{xy}{x^2+y^2}$ has no limit at the origin. Complex analysis inherits that difficulty and, remarkably, turns it into an advantage: the functions that survive the requirement are extremely well behaved.

Since $|z| \le |{\operatorname{Re}z}| + |{\operatorname{Im}z}|$ and $|\operatorname{Re}z|, |\operatorname{Im}z| \le |z|$, closeness in $\mathbb{C}$ and closeness in both coordinates are the same thing. So:

With $f = u+iv$ and $L = a+ib$,

$$\lim_{z\to z_0}f(z) = L \iff \lim_{(x,y)\to(x_0,y_0)}u = a \ \text{ and } \ \lim_{(x,y)\to(x_0,y_0)}v = b$$

This is how you import everything you know about multivariable limits. It also means continuity of $f$ is exactly continuity of $u$ and $v$ as functions of two real variables.

All the usual algebra survives, with the same proofs:

• $\lim(f\pm g) = \lim f \pm \lim g$
• $\lim(fg) = \lim f \cdot \lim g$
• $\lim(f/g) = \lim f / \lim g$ provided $\lim g \ne 0$
• Compositions of continuous functions are continuous

$f$ is continuous at $z_0$ if $z_0$ is in the domain and $\lim_{z\to z_0}f(z) = f(z_0)$. There is also a sequential criterion, exactly as in real analysis: $f$ is continuous at $z_0$ iff $f(z_n)\to f(z_0)$ for every sequence $z_n \to z_0$.

Continuous functions include all polynomials in $z$, all rational functions away from the zeros of the denominator, $e^z$, $\sin z$, $\cos z$, and also $\bar z$, $|z|$, $\operatorname{Re}z$, $\operatorname{Im}z$. That last group is the important observation: continuity is cheap, and none of those four is differentiable. The interesting line in complex analysis is drawn at differentiability, not continuity.

Because complex analysis works on the extended plane, three more limit statements are standard:

• $\lim_{z\to z_0}f(z) = \infty$ means $|f(z)|$ can be made arbitrarily large as $z\to z_0$. This is what having a pole at $z_0$ means.
• $\lim_{z\to\infty}f(z) = L$ means $f(z)$ approaches $L$ as $|z|$ grows without bound.
• $\lim_{z\to\infty}f(z) = \infty$, as for any non-constant polynomial.

A useful reduction: behaviour at $\infty$ is studied by substituting $z = 1/\zeta$ and looking at $\zeta \to 0$. Note there is only one $\infty$ in $\hat{\mathbb{C}}$, unlike the real line’s $\pm\infty$; on the Riemann sphere it is a single point, and every direction of escape heads for it.

Two words used constantly from here on:

• A set is open if every point has a disc around it inside the set, and connected if it is not split into two separated pieces. An open connected set is called a domain.
• A domain is simply connected if it has no holes, meaning every loop in it can be shrunk to a point without leaving it.

Simple connectivity is the hypothesis in Cauchy’s theorem, and the annulus is the standard example of a domain that fails it. That failure is exactly what Laurent series and residues exist to handle.

Example 1: Show $\lim_{z\to0}\frac{\bar z}{z}$ does not exist.

Solution. Put $z = te^{i\theta}$ with $t>0$:

$$\frac{\bar z}{z} = e^{-2i\theta}$$

Along $\theta = 0$ the value is 1 for every $t$, so that path gives 1.

Along $\theta = \pi/2$ the value is $e^{-i\pi} = -1$.

Two paths, two values, so the limit does not exist. ∎

The function is bounded and continuous away from 0, which is a good reminder that failure of a limit need not involve anything blowing up.

Example 2: Show $\lim_{z\to0}\frac{\operatorname{Re}(z)}{|z|}$ does not exist.

Solution. On the ray at angle $\theta$, $\operatorname{Re}z = t\cos\theta$ and $|z| = t$, so the quotient is $\cos\theta$.

Along the positive real axis it is 1; along the positive imaginary axis it is 0. ∎

Example 3: Evaluate $\lim_{z\to i}\frac{z^2+1}{z-i}$.

Solution. Factor, using $z^2+1 = (z-i)(z+i)$:

$$\frac{(z-i)(z+i)}{z-i} = z+i \quad (z \ne i)$$

So the limit is $i+i = 2i$. ∎

Removable singularities factor out, exactly as in real calculus. Nothing about direction independence causes trouble here, because after cancelling, the expression is a polynomial.

Example 4: Is $f(z)=\dfrac{z^2+1}{z-i}$ continuous at $i$?

Solution. No, because $f(i)$ is undefined - the point is not in the domain, so continuity is not even a meaningful question there.

However the limit exists and equals $2i$, so defining $f(i) = 2i$ makes the extended function continuous. This is a removable singularity, and it will turn out to be removable in a much stronger sense: the extended function is analytic, not merely continuous.

Example 5: Where is $f(z) = \dfrac{z}{z^2+4}$ continuous?

Solution. It is a rational function, so it is continuous wherever the denominator is nonzero. Solving $z^2+4=0$ gives $z = \pm2i$.

$$\text{continuous on } \mathbb{C}\setminus\{2i,\,-2i\}$$

At those two points $|f(z)|\to\infty$, so the function has poles there. ∎

Example 6: Verify continuity of $f(z) = z^2$ by splitting into $u$ and $v$.

Solution. We have $u(x,y) = x^2-y^2$ and $v(x,y) = 2xy$. Both are polynomials in two real variables, hence continuous everywhere on $\mathbb{R}^2$.

Since $f$ is continuous iff both $u$ and $v$ are, $f$ is continuous on all of $\mathbb{C}$. ∎

This is the standard route for transferring real results. Any statement about $f$ that is purely about closeness can be checked coordinatewise.

Example 7: An epsilon-delta proof.

Prove $\lim_{z\to z_0} z^2 = z_0^2$ directly.

Solution. Estimate the target quantity:

$$|z^2-z_0^2| = |z-z_0|\,|z+z_0|$$

Restrict attention to $|z-z_0|<1$, which forces $|z| < |z_0|+1$ and hence

$$|z+z_0| \le |z|+|z_0| < 2|z_0|+1$$

Given $\varepsilon>0$, choose

$$\delta = \min\left\{1,\ \frac{\varepsilon}{2|z_0|+1}\right\}$$

Then $|z-z_0|<\delta$ gives $|z^2-z_0^2| < \delta(2|z_0|+1) \le \varepsilon$. ∎

The proof is character-for-character the real one. That is worth noticing: the epsilon-delta machinery transfers unchanged, and the only new phenomenon is the extra freedom the approaching point has.

Numerical root-finding. Newton’s method in the complex plane converges to different roots depending on the starting point, and the boundaries between the basins are fractals. Whether the iteration has a limit, and which one, is exactly a question about limits of complex sequences.

Filter and system stability. Whether a transfer function stays bounded as $z\to\infty$ or blows up at a pole determines whether an amplifier saturates or a control loop runs away. Engineers read off pole locations precisely because those are the points where the limit is $\infty$.

Fractals. The Mandelbrot set is defined by whether the orbit of 0 under $z\mapsto z^2+c$ stays bounded. Every pixel is the answer to a boundedness question about a complex sequence.

Signal analysis near a resonance. The gain of a resonant system approaches a pole as the driving frequency approaches the natural frequency, and how fast it grows is set by the distance to that pole in the plane.

Analytic continuation in physics. Scattering amplitudes are continued from real energies into the complex plane, and the poles found there correspond to physical particles and resonances. The limiting behaviour on the way in is the whole content of the technique.

Computer graphics. Conformal image warps must be continuous to avoid visible tearing, and continuity is checked coordinatewise on $u$ and $v$ exactly as in Example 6.