[Conformal maps](/page/Conformal%20Maps) are the functions that let complex analysis behave like geometry. A holomorphic function can stretch, rotate, fold, or collapse small figures; the conformal condition isolates the case where, at each point, the first-order behaviour is only a rotation together with a positive change of scale. This is why conformal maps are central in complex analysis, [Holomorphic Function](/page/Holomorphic%20Function), [Derivative](/page/Derivative), and harmonic function theory: they transfer local analytic information while preserving oriented angles. The word "oriented" matters: anti-holomorphic maps preserve the size of angles but reverse their orientation, so they belong to a neighbouring convention rather than to the holomorphic one used here. The need for the nonvanishing derivative condition is already visible from the map $z \mapsto z^2$. Away from $0$ its first-order behaviour is multiplication by $2z$, so small angles are preserved at nonzero points. At $0$, the derivative vanishes and two distinct tangent directions can collapse into the same first-order direction. Conformal maps remove precisely this kind of critical behaviour. ## Definition Most applications need a map that has no critical points anywhere on its domain. Requiring a nonzero complex derivative at every point makes the map usable as a system of local complex coordinates throughout the open set, rather than merely as a differentiable function with good behaviour at isolated points. The domain is taken to be open because [complex differentiability](/page/Complex%20Differentiability) is defined through limits from nearby points in every direction. [definition: Conformal Map] Let $U \subset \mathbb C$ be an open set. A function $f: U \to \mathbb C$ is a conformal map on $U$ if $f$ is holomorphic on $U$ and \begin{align*} f'(w) \ne 0 \end{align*} for every $w \in U$. [/definition] The global definition is built from a local test. Isolating the pointwise condition is useful because failures of conformality usually occur at critical points, and examples such as $z \mapsto z^2$ fail at a single point while remaining conformal elsewhere. [definition: Conformal at a Point] Let $U \subset \mathbb C$ be an open set, let $w \in U$, and let $f: U \to \mathbb C$ be holomorphic on a neighbourhood of $w$. The map $f$ is conformal at $w$ if \begin{align*} f'(w) \ne 0. \end{align*} [/definition] A conformal map need not be globally injective. The exponential function on a horizontal strip of width greater than $2\pi$ is locally conformal everywhere but identifies points differing by $2\pi i$. Because global one-to-one behaviour is often the relevant notion in classifying domains, it deserves its own definition. [definition: Conformal Isomorphism] Let $U,V \subset \mathbb C$ be open sets. A conformal isomorphism from $U$ to $V$ is a bijective conformal map $f: U \to V$. [/definition] The definition of conformal isomorphism would be too weak if it only described the forward map. In applications, a conformal change of coordinates is meant to be reversible: a function, contour, or boundary value problem should be transported to the target domain and then brought back without leaving holomorphic geometry. Bijectivity alone gives a set-theoretic inverse, but the analytic content is that this inverse is holomorphic and has no critical points. [quotetheorem:8269] This closure under inverses is why conformal isomorphisms are the correct equivalences in planar complex geometry. They identify domains that have the same complex-analytic structure even when their Euclidean shapes look different. Conformality preserves angles but not lengths. To measure the allowed length distortion, one separates the derivative into a rotational part and a positive size. The positive size is the quantity that appears when arc length, area, or the Laplacian is transformed. [definition: Conformal Scale Factor] Let $U \subset \mathbb C$ be an open set and let $f: U \to \mathbb C$ be conformal on $U$. The conformal scale factor of $f$ is the function $\lambda_f: U \to (0,\infty)$ defined by \begin{align*} \lambda_f(w) = |f'(w)|. \end{align*} [/definition] The scale factor records metric distortion, while the argument of $f'(w)$ records rotation. Together they recover the first-order action of $f$ near $w$ as multiplication by the complex number $f'(w)$. For a compact first model, take the maps whose derivative is the same nonzero complex number everywhere. They show the definition with no hidden global complication. [example: Nonzero Complex Affine Map] Let $a,b\in\mathbb C$ with $a\ne 0$, and define $f:\mathbb C\to\mathbb C$ by $f(z)=az+b$. For any $z\in\mathbb C$ and any nonzero $h\in\mathbb C$, \begin{align*} \frac{f(z+h)-f(z)}{h}=\frac{a(z+h)+b-(az+b)}{h}=\frac{ah}{h}=a. \end{align*} Hence the limit as $h\to 0$ is $a$, so $f'(z)=a$ for every $z\in\mathbb C$. Since $a\ne 0$, the derivative is nonzero at every point, and therefore $f$ is conformal on $\mathbb C$. Its conformal scale factor is \begin{align*} \lambda_f(z)=|f'(z)|=|a|, \end{align*} so the scale is constant. Multiplication by $a$ rotates tangent directions by $\arg(a)$ and stretches their lengths by $|a|$. To see that $f$ is a conformal isomorphism, define $g:\mathbb C\to\mathbb C$ by $g(w)=(w-b)/a$. Then \begin{align*} g(f(z))=\frac{az+b-b}{a}=z \end{align*} for every $z\in\mathbb C$, and \begin{align*} f(g(w))=a\frac{w-b}{a}+b=w-b+b=w \end{align*} for every $w\in\mathbb C$. Thus $g=f^{-1}$, so $f$ is bijective; because $f$ is conformal on $\mathbb C$, it is a conformal isomorphism from $\mathbb C$ to $\mathbb C$. [/example] This example is the local model for every conformal map: near each point, a conformal map behaves to first order like a nonzero affine map. Nonlinear conformal maps differ because their scale and rotation can vary from point to point. ## Equivalent Characterisations The definition is short because complex differentiability is already a strong condition. To see the geometry inside it, write a complex function as a map between real planes. If $f(z) = u(x,y) + i v(x,y)$ with $z = x + iy$, then the real derivative at a point is a [linear map](/page/Linear%20Map) from $\mathbb R^2$ to $\mathbb R^2$. [definition: Real Jacobian of a Complex Function] Let $U \subset \mathbb C$ be open, let $f: U \to \mathbb C$ be written as $f(x+iy)=u(x,y)+iv(x,y)$, and let $w=x+iy \in U$. If the first partial derivatives exist at $w$, the real Jacobian matrix of $f$ at $w$ is the matrix $Jf_w \in \mathbb R^{2\times 2}$ with entries $(Jf_w)_{11}=\partial_xu(w)$, $(Jf_w)_{12}=\partial_yu(w)$, $(Jf_w)_{21}=\partial_xv(w)$, and $(Jf_w)_{22}=\partial_yv(w)$. [/definition] The real Jacobian is the object that acts on tangent vectors in the plane. To connect the analytic definition with planar geometry, we need to know exactly what the [Cauchy-Riemann equations](/page/Cauchy-Riemann%20Equations) do to this matrix and when that matrix is invertible. [quotetheorem:8270] This theorem explains why conformal maps preserve orientation: the determinant is positive. Anti-holomorphic maps such as $z \mapsto \bar z$ preserve the size of angles but reverse orientation, so they are not conformal under the holomorphic convention used here. The matrix criterion still leaves a geometric question unresolved. A derivative acts first on infinitesimal displacement vectors, but the visible objects in a domain are curves meeting at angles. To justify calling the map conformal, we need the nonzero complex derivative to control every tangent direction in the same way: it should rotate all directions by one angle and scale all lengths by one positive factor, so the angle between any two smooth curves through the point is unchanged. [quotetheorem:334] The derivative condition is therefore not arbitrary. It encodes the geometric request that all tangent directions at a point be rotated by the same amount and stretched by the same positive factor. ## Standard Examples ### Covering-Like Local Maps The exponential function illustrates the distinction between local conformality and global injectivity. It has no critical points, but its periodicity prevents it from being one-to-one on large domains. [example: Exponential Map on a Strip] Let \begin{align*} S=\{z\in\mathbb C:0<\operatorname{Im}(z)<\pi\}, \end{align*} and define $f:S\to\mathbb C$ by $f(z)=e^z$. The complex exponential is holomorphic and satisfies \begin{align*} f'(z)=e^z. \end{align*} Since $e^z\ne 0$ for every $z\in\mathbb C$, we have $f'(z)\ne 0$ for every $z\in S$, so $f$ is conformal on $S$. We compute the image. Write $z=x+iy$ with $x,y\in\mathbb R$ and $0<y<\pi$. Then \begin{align*} e^z=e^{x+iy}=e^x e^{iy}=e^x(\cos y+i\sin y). \end{align*} Thus \begin{align*} \operatorname{Im}(e^z)=e^x\sin y. \end{align*} Here $e^x>0$ and $\sin y>0$ for $0<y<\pi$, so $\operatorname{Im}(e^z)>0$. Hence $f(S)\subseteq\{w\in\mathbb C:\operatorname{Im}(w)>0\}$. Conversely, let $w$ lie in the upper half-plane. Then $w=re^{i\theta}$ for some $r>0$ and some $\theta\in(0,\pi)$. Put \begin{align*} z=\log r+i\theta. \end{align*} Then $z\in S$, and \begin{align*} e^z=e^{\log r+i\theta}=e^{\log r}e^{i\theta}=re^{i\theta}=w. \end{align*} Therefore \begin{align*} f(S)=\{w\in\mathbb C:\operatorname{Im}(w)>0\}. \end{align*} It remains to check injectivity on this strip. Suppose $z_1=x_1+iy_1$ and $z_2=x_2+iy_2$ belong to $S$ and $e^{z_1}=e^{z_2}$. Taking absolute values gives \begin{align*} e^{x_1}=|e^{z_1}|=|e^{z_2}|=e^{x_2}. \end{align*} Since the real exponential is injective, $x_1=x_2$. Dividing the equality $e^{z_1}=e^{z_2}$ by the common positive factor $e^{x_1}=e^{x_2}$ gives \begin{align*} \cos y_1+i\sin y_1=\cos y_2+i\sin y_2. \end{align*} Thus $\cos y_1=\cos y_2$ and $\sin y_1=\sin y_2$. Since both $y_1$ and $y_2$ lie in $(0,\pi)$, the angle in this interval is determined by its sine and cosine, so $y_1=y_2$. Hence $z_1=z_2$, and $f$ is injective on $S$. The exponential map is therefore a bijective conformal map from $S$ onto the upper half-plane, so it is a conformal isomorphism from the strip $S$ to $\{w\in\mathbb C:\operatorname{Im}(w)>0\}$. [/example] This example is a standard bridge between rectangular or strip-shaped domains and half-plane geometry. It also foreshadows how conformal maps can simplify boundary value problems by moving them to a more convenient domain. ### Critical Points and Local Failure A holomorphic function can fail to be conformal at isolated points even when it is nonconstant and locally well behaved elsewhere. Critical points are the basic obstruction. [example: Squaring Map and a Critical Point] Define $f:\mathbb C\to\mathbb C$ by $f(z)=z^2$. For $h\ne 0$, \begin{align*} \frac{f(z+h)-f(z)}{h}=\frac{(z+h)^2-z^2}{h}. \end{align*} Expanding the square gives \begin{align*} (z+h)^2-z^2=z^2+2zh+h^2-z^2=2zh+h^2. \end{align*} Therefore \begin{align*} \frac{f(z+h)-f(z)}{h}=\frac{2zh+h^2}{h}=2z+h. \end{align*} Taking the limit as $h\to 0$ gives \begin{align*} f'(z)=2z. \end{align*} If $z\ne 0$, then $2z\ne 0$, so $f$ is conformal at every point of $\mathbb C\setminus\{0\}$. At $z=0$, however, \begin{align*} f'(0)=2\cdot 0=0, \end{align*} so $f$ is not conformal at $0$. The angle failure can be seen from two rays through $0$. Let $\gamma_1(t)=t$ and $\gamma_2(t)=it$. Then \begin{align*} \gamma_1'(0)=1 \end{align*} and \begin{align*} \gamma_2'(0)=i, \end{align*} so their tangent directions meet at oriented angle $\pi/2$. Their images under $f$ are \begin{align*} (f\circ\gamma_1)(t)=f(t)=t^2 \end{align*} and \begin{align*} (f\circ\gamma_2)(t)=f(it)=(it)^2=i^2t^2=-t^2. \end{align*} Differentiating these real-parameter curves gives \begin{align*} (f\circ\gamma_1)'(t)=2t \end{align*} and \begin{align*} (f\circ\gamma_2)'(t)=-2t. \end{align*} At $t=0$ both image tangents vanish: \begin{align*} (f\circ\gamma_1)'(0)=0 \end{align*} and \begin{align*} (f\circ\gamma_2)'(0)=0. \end{align*} Thus the two nonzero tangent directions at the source no longer determine nonzero first-order tangent directions after applying $f$. The failure of conformality is exactly the critical point where $f'(z)=2z$ vanishes. [/example] This failure example explains why the derivative must be nonzero at every point, not merely at most points. A single critical point changes the local geometry and prevents the map from having a holomorphic local inverse there. ### Punctured-Domain Symmetries The reciprocal map gives a conformal transformation of a punctured domain and shows that conformality naturally lives on open sets that need not be simply connected. [example: Reciprocal Map on the Punctured Plane] Let $U=\mathbb C\setminus\{0\}$ and define $f:U\to U$ by \begin{align*} f(z)=\frac{1}{z}. \end{align*} This is well-defined because if $z\in U$, then $z\ne 0$, so $1/z$ exists and $1/z\ne 0$. We compute the derivative at a fixed $z\in U$. For $h\ne 0$ with $z+h\ne 0$, \begin{align*} \frac{f(z+h)-f(z)}{h}=\frac{\frac{1}{z+h}-\frac{1}{z}}{h}. \end{align*} Putting the two fractions over the common denominator $z(z+h)$ gives \begin{align*} \frac{1}{z+h}-\frac{1}{z}=\frac{z-(z+h)}{z(z+h)}. \end{align*} Since $z-(z+h)=-h$, we get \begin{align*} \frac{f(z+h)-f(z)}{h}=\frac{-h}{h z(z+h)}. \end{align*} Cancelling the nonzero factor $h$ gives \begin{align*} \frac{f(z+h)-f(z)}{h}=-\frac{1}{z(z+h)}. \end{align*} Taking $h\to 0$ yields \begin{align*} f'(z)=-\frac{1}{z^2}. \end{align*} Because $z\ne 0$, we have $z^2\ne 0$, so $-1/z^2\ne 0$. Thus $f'(z)\ne 0$ for every $z\in U$, and $f$ is conformal on $U$. It remains to check that $f$ is a bijection from $U$ to itself. For every $z\in U$, \begin{align*} f(f(z))=f\left(\frac{1}{z}\right)=\frac{1}{1/z}=z. \end{align*} Hence each $w\in U$ has a preimage, namely $f(w)$, because \begin{align*} f(f(w))=w. \end{align*} So $f$ is surjective. If $f(z_1)=f(z_2)$, then applying $f$ to both sides gives \begin{align*} f(f(z_1))=f(f(z_2)). \end{align*} Using the identity just proved, this becomes \begin{align*} z_1=z_2. \end{align*} So $f$ is injective. Therefore the reciprocal map is a conformal isomorphism from the punctured plane $U$ to itself. [/example] The reciprocal map reverses radial size in the sense that points near $0$ are sent far away, but it still preserves oriented angles locally because its derivative is nonzero complex multiplication at each point. ## Local Inverses and Analytic Transport The most important local consequence of conformality is that a conformal map can be inverted near each point. This turns the nonvanishing derivative condition into a practical tool for changing coordinates. [quotetheorem:4950] This theorem is the point where the local nonzero-derivative test becomes a usable coordinate theorem. It does not say that a conformal map is globally one-to-one, and examples such as the exponential map on a wide strip show why that would be false. Its content is local: near any point where the derivative is nonzero, the map has a holomorphic inverse with nonzero derivative, so small pieces of the domain can be moved back and forth without leaving the category of conformal maps. That local reversibility is the reason conformal maps can be treated as changes of complex coordinates, and it is the mechanism behind inverse maps of conformal isomorphisms. Conformal maps should be stable under successive changes of variable. If $f$ carries $U$ into $V$ without critical points and $g$ carries $V$ into $W$ without critical points, the natural question is whether $g\circ f$ still has no critical points. The chain rule answers that question. [quotetheorem:8271] The point of the composition theorem is that conformal changes of variable can be chained without leaving the same class of maps. This is what makes conformal equivalence behave like a usable geometry rather than a one-step construction: a map from $U$ to $V$ and then from $V$ to $W$ gives a conformal route from $U$ to $W$. The derivative formula also records exactly how local scale and rotation accumulate. The scale factors multiply as $|g'(f(z))|\,|f'(z)|$, so no new critical point appears in the composition as long as neither factor has one. The next reason conformal maps matter is not just geometric but analytic. Harmonic functions are governed by the Laplacian, and in two real dimensions the Laplacian transforms especially well under a conformal change of variables. Here $C^2(V;\mathbb R)$ denotes the twice continuously differentiable real-valued functions on $V$, and $\Delta u=0$ means that the Laplacian of $u$ vanishes. [quotetheorem:8272] This pullback theorem explains why conformal maps are useful in potential theory: they let harmonic functions be transported from a target domain back to a source domain without destroying harmonicity. The result is one-way as stated, because it pulls a function on $V$ back along a conformal map $f:U\to V$; when $f$ is a conformal isomorphism, the inverse theorem lets the same idea be applied in both directions. In practice this is how a boundary value problem on a complicated domain can be moved to a simpler one, solved there, and then interpreted back on the original domain. Classification questions require a way to say that two domains have the same complex-analytic structure. The formal language for this sameness is an [equivalence relation](/page/Equivalence%20Relation), which lets classification problems be phrased without choosing a privileged representative. [quotetheorem:8273] This equivalence relation is the organizing language behind classification questions. Instead of asking whether two domains look the same as subsets of the plane, conformal geometry asks whether their holomorphic function theories are the same after a conformal change of variable. ## Beyond and Connected Topics Conformal maps refine [Continuity](/page/Continuity), differentiability, and holomorphicity. A continuous map can distort oriented angles without restriction; a real differentiable map can preserve oriented angles at one point without satisfying any complex-analytic constraints nearby. Holomorphicity supplies rigidity, while the nonzero derivative condition supplies nondegeneracy. The connection with [Open Set](/page/Open%20Set) is structural rather than cosmetic. Domains of holomorphic functions must be open, and the local inverse theorem produces open image neighbourhoods. The [open mapping theorem](/theorems/631) also shows that every nonconstant holomorphic function maps open sets to open sets, but conformal maps have the stronger local coordinate property given by nonvanishing derivative. Conformal maps are also related to [Topology](/page/Topology). Since a conformal isomorphism is defined analytically, it is useful to separate what part of its behaviour is already topological: continuity comes from holomorphicity, and continuity of the inverse comes from the inverse theorem. [quotetheorem:8274] Thus conformal isomorphisms preserve topological features such as connectedness and holes. They also preserve analytic features such as holomorphic functions after composition, which is why conformal equivalence is finer than mere homeomorphism. On a Riemann surface, conformal maps become holomorphic coordinate changes with nonzero derivative in local charts. This viewpoint turns the plane-level definition into the local model for complex one-dimensional geometry. A final warning is worth isolating: preservation of angle size alone is not enough to identify a conformal map unless orientation and regularity are specified. The conjugation map $z\mapsto \bar z$ preserves the size of Euclidean angles but reverses orientation and is anti-holomorphic rather than holomorphic. The holomorphic convention fixes the orientation-preserving theory. [remark: Orientation Convention] In this page, conformal means holomorphic with nonzero complex derivative. Some texts use conformal for angle-preserving maps in a broader sense and then distinguish holomorphic conformal maps from anti-holomorphic conformal maps. Androma uses the holomorphic convention for this concept page. [/remark] ## References Lars Ahlfors, *Complex Analysis* (1979). John B. Conway, *Functions of One Complex Variable I* (1978). Theodore W. Gamelin, *Complex Analysis* (2001). [Holomorphic Function](/page/Holomorphic%20Function). [Derivative](/page/Derivative).