Lesson 8: Conformal Mapping & Möbius Transformations

Every lesson so far has asked what a complex function computes. This one asks what it does—where it sends points, what shapes it turns into what other shapes. The answer starts with a single unremarkable-looking function, $1/z$, and ends with a theorem saying that almost every region in the plane is secretly the unit disk.

What $1/z$ actually does

Before any definitions, compute. Take $T(z)=1/z$.

On the unit circle. $T(1)=1$, $T(i)=-i$, $T(-1)=-1$, $T(-i)=i$, and generally $T(e^{i\theta})=e^{-i\theta}$, straight from Euler's identity. The unit circle maps to itself, with each point $e^{i\theta}$ landing at $e^{-i\theta}$—its mirror image across the real axis.

It's worth being precise about what that does and doesn't mean. It is not true that $T(z)=-z$ in general (check $T(1)=1$, not $-1$). What gets negated is the angle, not the value. There is a clean reason the image looks like a reflection here specifically: for any $z\ne0$,

$$\frac1z=\frac{\bar z}{z\bar z}=\frac{\bar z}{|z|^2}$$

and on the unit circle $|z|=1$, so this collapses to $1/z=\bar z$ exactly. Away from $|z|=1$ that fails. It's a coincidence of sitting on the unit circle, not a general property of inversion.

Does the backwards traversal break conformality? No, and the reason matters. $T'(z)=-1/z^2$, so $T'(1)=-1$: a rotation by $\arg(-1)=\pi$ with scale factor $1$. A rotation preserves both the size and the sense of angles between curves, so $T$ is conformal at $z=1$, full stop. The apparent backwards sweep is a global fact about that one curve mapping onto itself, not a local orientation flip. A genuinely orientation-reversing map would be $z\mapsto\bar z$, which isn't even holomorphic. It fails the Cauchy–Riemann equations, so no complex derivative exists. Meanwhile $T(z)=1/z$ is holomorphic with $T'(z)\ne0$ everywhere on its domain, hence a rotation-plus-scaling at every point, never a reflection.

On the line $\text{Re}(z)=1$. Computing $T(1+yi)=\frac1{1+yi}$ for $y=-1,0,1,2$ gives $0.5+0.5i$, $1$, $0.5-0.5i$, $0.2-0.4i$, all four sitting at distance exactly $0.5$ from the point $0.5$ on the real axis. The line becomes a circle, and one passing through the origin.

That single computation is the whole lesson in miniature. A Möbius transformation is a map

$$T(z)=\frac{az+b}{cz+d},\qquad a,b,c,d\in\mathbb{C},\quad ad-bc\ne0$$

(the determinant condition rules out the degenerate case where $T$ collapses to a constant). Inversion $1/z$ is the case $a=0$, $b=1$, $c=1$, $d=0$.

Key takeaway: $1/z$ sends the unit circle to itself by conjugating angles, and sends the line $\text{Re}(z)=1$ to a circle through the origin. It is conformal everywhere on its domain, since $T'(z)=-1/z^2$ is never zero.

Why circles and lines always map to circles and lines

A little vocabulary first, since the argument leans on it. A map is just another word for function, used when the emphasis is geometric (where does each point go?) rather than computational. A linear map is $z\mapsto\lambda z$: a uniform rotation-and-scaling by the constant $\lambda$, exactly Lesson 2's picture of complex multiplication. (Take $\lambda=2i$: scale by $2$, rotate by $90°$.) An affine map is $z\mapsto\lambda z+\mu$: a linear map followed by a rigid translation. By contrast $z\mapsto z^2$ is not affine: it acts differently depending on where $z$ sits, whereas an affine map performs the identical rigid transformation everywhere. That distinction is precisely why affine maps preserve circles and lines for free, and why $z^2$ and $1/z$ don't get to claim it automatically.

The decomposition. For $c\ne0$, polynomial division gives

$$T(z)=\frac{az+b}{cz+d}=\frac ac+\frac{bc-ad}{c}\cdot\frac1{cz+d}$$

So every Möbius transformation is built from three pieces in sequence: an affine map $z\mapsto cz+d$, then inversion $w\mapsto1/w$, then another affine map (scale by $\frac{bc-ad}{c}$, translate by $\frac ac$). Since affine maps can't bend a line into a curve, the entire question collapses to a single case: does inversion preserve circles and lines?

The general proof. Every circle or line in the plane is

$$A(x^2+y^2)+Bx+Cy+D=0$$

with $A\ne0$ giving a circle and $A=0$ giving a line. Substitute $x=\frac{z+\bar z}2$, $y=\frac{z-\bar z}{2i}$, and $x^2+y^2=z\bar z$. Simplifying $\frac{C}{2i}=-\frac{iC}2$ (multiply through by $\frac{-i}{-i}$), the equation becomes

$$Az\bar z+Ez+\bar E\bar z+D=0,\qquad E:=\frac{B-iC}{2}$$

Now substitute $w=1/z$, so $z=1/w$. Since conjugation and reciprocation commute, $\overline{1/w}=1/\bar w$, giving $\bar z=1/\bar w$ as well. Clearing denominators by multiplying through by $w\bar w$:

$$Dw\bar w+\bar Ew+E\bar w+A=0$$

The same shape as the original, with $A$ and $D$ swapped (and $E\to\bar E$). So inversion sends every circle-or-line to a circle-or-line, and combined with the decomposition above: every Möbius transformation sends circles and lines to circles and lines. $\blacksquare$

Reading the swap geometrically. The algebra is clean, but translating it back into "which shapes become which" is the step worth slowing down on. For the line $\text{Re}(z)=1$ we have $A=0$, $B=1$, $C=0$, $D=-1$. Here $A=0$ says it's a line, and $D\ne0$ says it misses the origin. After the swap, the new leading coefficient is the old $D=-1\ne0$ (a genuine circle), and the new constant term is the old $A=0$, meaning the new circle passes through the origin. Which is exactly what the direct computation found: center $0.5$, radius $0.5$, threaded through $0$.

So in general: any line missing the origin becomes, under inversion, a circle through the origin. Not a quirk of one example—precisely what the $A\leftrightarrow D$ swap asserts every time.

Two more cases make the pattern concrete. The line $\text{Re}(z)=2$ also misses the origin, so it too becomes a circle through the origin, just at a different offset. The line $y=x$ passes through the origin, so $A=0$ and $D=0$ both hold; after the swap both stay zero, and the image is still a line through the origin. Worth one extra step of care, though: parametrizing $y=x$ as $z=t(1+i)$ gives $\frac1z=\frac{1-i}{2t}$, which traces $y=-x$. The type is preserved exactly as predicted, but the specific line is the mirror image, not the original.

Drag the source curve below and watch its image redraw under $w=1/z$. The one interaction worth doing deliberately: drag a line until it crosses the origin, and watch its image snap from a circle back to a line at the instant $D$ hits zero. The coefficient equation sits below the panels, where $A$ and $D$ visibly trade places as the shape changes type.

Key takeaway: write any circle-or-line as $Az\bar z+Ez+\bar E\bar z+D=0$; inversion swaps $A\leftrightarrow D$ and conjugates $E$. Since Möbius maps decompose into affine–inversion–affine, they all send circles and lines to circles and lines. Reading the swap: $A=0$ means line, $D=0$ means through the origin, so a line missing the origin becomes a circle through it.

Möbius maps are conformal everywhere

Lesson 2 established that a holomorphic map is conformal wherever $f'(z_0)\ne0$. Differentiating by the quotient rule:

$$T'(z)=\frac{a(cz+d)-c(az+b)}{(cz+d)^2}=\frac{ad-bc}{(cz+d)^2}$$

The nondegeneracy condition now does double duty. We already needed $ad-bc\ne0$ to keep $T$ from collapsing to a constant; that same condition guarantees the numerator of $T'$ is never zero. The denominator $(cz+d)^2$ vanishes only at $z=-d/c$, which is already excluded from $T$'s domain as its pole.

So $T'(z)\ne0$ everywhere $T$ is defined. Möbius maps are conformal across their entire domain, with no exceptional points to hunt for—unlike a generic holomorphic function such as $z^2$, whose conformality fails at the isolated point where its derivative vanishes.

Key takeaway: $T'(z)=\dfrac{ad-bc}{(cz+d)^2}$. The single condition $ad-bc\ne0$ makes $T$ both non-constant and conformal everywhere at once.

Group structure, and three points determining a map

Möbius transformations form a group

Closure, concretely: with $T_1(z)=\frac1z$ and $T_2(z)=\frac{z+1}{z-1}$,

$$(T_2\circ T_1)(z)=T_2\!\left(\tfrac1z\right)=\frac{1+z}{1-z}$$

That's Möbius again. In general, composing two linear-over-linear fractions and clearing denominators always produces another one. (This is $2\times2$ matrix multiplication in disguise, which is a genuinely useful way to think about it, though nothing below depends on it.)

Identity: $T(z)=z$, the case $a=1,b=0,c=0,d=1$.

Inverse: solve $w=\frac{az+b}{cz+d}$ for $z$. From $w(cz+d)=az+b$ we get $z(wc-a)=b-wd$, so

$$z=\frac{-dw+b}{wc-a}$$

Also Möbius, and its determinant is $(-d)(-a)-bc=ad-bc$, the same nonzero quantity as the original, so nondegeneracy transfers for free.

Associativity is free, since function composition always associates. All four axioms hold: the Möbius transformations form a group under composition.

Three distinct points determine the map

Why three and not four. Scaling all four coefficients, $(a,b,c,d)\to(\lambda a,\lambda b,\lambda c,\lambda d)$ for any $\lambda\ne0$, leaves $T$ completely unchanged—the $\lambda$ cancels between numerator and denominator. So the four letters encode only three independent parameters, which is exactly how many point-conditions we should expect to be able to impose.

The tool. For distinct $z_1,z_2,z_3$, define

$$T(z)=\frac{(z-z_1)(z_2-z_3)}{(z-z_3)(z_2-z_1)}$$

Then $T(z_1)=0$, since $(z-z_1)$ kills the numerator, and $T(z_3)=\infty$, since $(z-z_3)$ kills the denominator. The third value is the one worth checking carefully rather than guessing: at $z=z_2$ the numerator and denominator turn out to contain the same product $(z_2-z_1)(z_2-z_3)$, so $T(z_2)=1$, not $0$.

Existence. To send arbitrary $z_1,z_2,z_3$ to arbitrary $w_1,w_2,w_3$: build $T$ sending the $z_i\to0,1,\infty$ and build $S$ sending the $w_i\to0,1,\infty$ the same way. Then $S^{-1}\circ T$ does the job, using the group structure just established. Concretely, with $z_1,z_2,z_3=0,1,2$ and $w_1,w_2,w_3=1,0,-1$:

$$T(z)=\frac{-z}{z-2},\qquad S(w)=\frac{1-w}{1+w},\qquad (S^{-1}\circ T)(z)=1-z$$

($S$ happens to be self-inverse here.) Verified at all three points, including $z_3=2\mapsto1-2=-1=w_3$.

Uniqueness, by counting fixed points. Suppose $T_1$ and $T_2$ both send $z_1,z_2,z_3$ to $w_1,w_2,w_3$. Then $T_2^{-1}\circ T_1$ fixes all three $z_i$. That restatement is the crux of the argument and worth stating separately from what preceded it: "$T_1$ and $T_2$ agree at $z_i$" and "$T_2^{-1}\circ T_1$ fixes $z_i$" are two different sentences, and only the second one lets us count.

Now count. Setting $T(z)=z$ for a general Möbius map and clearing denominators gives

$$cz^2+(d-a)z-b=0$$

a quadratic with at most $2$ roots, unless $c=0$, in which case it's linear with at most $1$ root, unless every coefficient vanishes, which forces $T=\text{id}$. So a non-identity Möbius map has at most two fixed points. Three distinct fixed points therefore force $T_2^{-1}\circ T_1=\text{id}$, i.e. $T_1=T_2$.

Theorem. Given distinct $z_1,z_2,z_3$ and distinct $w_1,w_2,w_3$, there is a unique Möbius map with $T(z_i)=w_i$.

There's a way to see this whole group at once. Add the point $\infty$ to the plane and the result is a sphere—the Riemann sphere—via stereographic projection from the north pole: the origin sits at the south pole, the unit circle becomes the equator, and $\infty$ is the north pole itself. On the sphere the group structure turns geometric. Grab the sphere below and spin it: every rigid rotation projects down to a Möbius map of the plane. Spinning about the polar axis gives $z\mapsto e^{i\theta}z$; flipping it $180°$ about the axis through $\pm1$ gives $z\mapsto 1/z$, swapping the two poles $0$ and $\infty$. And because a straight line lifts to a circle through the north pole while an ordinary circle doesn't, moving the sphere makes lines and circles trade places: the Segment 1 theorem, now visually inevitable, since on the sphere they are all simply circles.

Not every Möbius map is a rigid rotation, though, and the visualization is honest about it: scalings $z\mapsto\lambda z$ and translations $z\mapsto z+b$ act on the same sphere as non-rigid flows, sliding points across the surface rather than turning the ball. The rotations are exactly the Möbius maps that are rigid motions of the sphere, a proper subgroup of the whole group. This is also the cleanest way to picture the Riemann Mapping Theorem's uniqueness clause in the next section: the disk's own symmetries are Möbius maps, and "unique up to a disk automorphism" is precisely the freedom to compose with one of them.

Key takeaway: Möbius maps form a group under composition. The canonical map $T(z)=\frac{(z-z_1)(z_2-z_3)}{(z-z_3)(z_2-z_1)}$ sends $z_1,z_2,z_3\to0,1,\infty$; composing two of these sends any three points to any other three. Uniqueness follows because a non-identity Möbius map has at most two fixed points.

The Riemann Mapping Theorem

A region is simply connected if it is connected and has no holes—every loop inside it can be shrunk continuously to a point without leaving the region.

Theorem. For any nonempty, simply connected, proper subset $\Omega\subsetneq\mathbb{C}$, there is a biholomorphic map $f:\Omega\to\mathbb{D}$ onto the open unit disk. Fixing a basepoint $z_0\in\Omega$ and requiring $f(z_0)=0$ and $f'(z_0)>0$ makes $f$ unique.

This is a genuinely startling statement. $\Omega$ can be jagged, elongated, slit, or otherwise nothing like a disk, and a conformal map still identifies it with $\mathbb{D}$, because a conformal map only sees angles and local structure, never global shape. Every tool built up over the previous seven lessons (Cauchy's formula, power series, the identity theorem) transfers between $\Omega$ and $\mathbb{D}$ along $f$. In that precise sense the two regions are conformally indistinguishable: no holomorphic invariant can tell them apart.

The visualization below shows this directly. Both presets are genuine closed-form conformal maps of the disk (the deformed region really is $\mathbb{D}$, not an approximation of one), and the probe's two arms stay exactly perpendicular in both panels throughout the morph, since the map's derivative is a rotation-and-scaling.

Why $\mathbb{C}$ itself is excluded. A tempting first guess is that an unbounded domain simply can't map onto a bounded target. The half-plane example below refutes that immediately: $\mathbb{H}$ is unbounded, and it works fine. The real obstruction is narrower and more interesting.

Suppose $f:\mathbb{C}\to\mathbb{D}$ were a biholomorphism. Its domain being all of $\mathbb{C}$ makes $f$ entire by definition, and its range being confined to $\mathbb{D}$ makes it bounded, regardless of how large the domain is. Entire plus bounded forces $f$ constant, by Liouville's theorem (proved back in Lesson 5), and a constant map cannot be a bijection. Contradiction.

A map defined only on a proper subset such as $\mathbb{H}$ is never entire in the first place, so Liouville's hypothesis never triggers. The dividing line is "$\Omega=\mathbb{C}$ or not," not "bounded or not", which is why $\Omega\subsetneq\mathbb{C}$ is a real hypothesis of the theorem rather than a technicality. Switching the visualization to its $\Omega=\mathbb{C}$ preset shows the failure directly: the grid escapes the disk no matter how hard the attempted map compresses it.

For nice domains, the map can be Möbius. Everything in this lesson converges here. The map

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

sends the upper half-plane $\mathbb{H}$ onto $\mathbb{D}$, and it is nothing more exotic than a Möbius transformation. Check it: $f(i)=0$, so the interior point $i$ lands at the disk's center; and $f(0)=-1$, so the boundary point $0$ (the real axis being $\partial\mathbb{H}$) lands on $\partial\mathbb{D}$. Those are consistent rather than contradictory, since neither point lies inside its respective open set.

One honest caveat. The general proof of the Riemann Mapping Theorem is non-constructive. It establishes existence and uniqueness through normal families and Montel's theorem, and hands back no formula whatsoever. The maps in this lesson, the Möbius map above and the two families in the visualization, are known closed-form conformal maps chosen precisely so that what's shown is a real Riemann map rather than a plausible-looking interpolation. For a general region $\Omega$, the theorem promises the map exists without telling you how to write it down.

Key takeaway: every nonempty, simply connected $\Omega\subsetneq\mathbb{C}$ is biholomorphic to $\mathbb{D}$, uniquely once normalized by $f(z_0)=0$, $f'(z_0)>0$. $\mathbb{C}$ is excluded because a biholomorphism $\mathbb{C}\to\mathbb{D}$ would be entire and bounded, hence constant by Liouville—the obstruction is being all of $\mathbb{C}$, not being unbounded. For $\mathbb{H}$ the map is Möbius: $\frac{z-i}{z+i}$. The general theorem is non-constructive.