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At a vertex with exterior turning angle ${\beta }_j\pi$, the interior angle is $(1-{\beta }_j)\pi$. A local conformal map from a half-plane therefore behaves as $f(Z)-z_j\sim c_j(Z-X_j{)}^{1-{\beta }_j}$, so its derivative has local exponent $-{\beta }_j$.

On each real boundary interval between prevertices, $\arg f^{\prime }$ is constant because the image is a straight side. Across a prevertex its change is the exterior turning angle. Taking $\arg (Z-X_j)\in (0,\pi )$, the product ${\prod }_j(Z-X_j{)}^{-{\beta }_j}$ has exactly these changes. Dividing $f^{\prime }$ by that product removes the corner singularities and leaves a holomorphic nonzero function with constant boundary argument. Schwarz reflection and the regular behavior at infinity make it constant. This gives the Schwarz-Christoffel formulaThere are three freely normalizable real prevertices, because an automorphism of the upper half-plane has three real parameters. After those are fixed, the remaining $n-3$ prevertex parameters are determined by the polygon's shape through the side-length conditions. They are not further independent normalization choices.

If the last prevertex is at infinity, absorb its constant scale into $C$ and omit its finite factor:For a bounded polygon, ${\sum }_j{\beta }_j=2$, so this derivative behaves like $Z^{{\beta }_n-2}$, consistent with the corner described in the local coordinate $1/Z$.

The conformal map $\zeta =z^3$ sends the wedge to the upper half-plane. A fluid vortex at $c^3$ there has an opposite image at ${\bar{c}}^3$, so the pulled-back potential isThe two boundary rays map to the real axis, where the imaginary part is constant, giving the required impermeability. Factoring $z^3-c^3=(z-c)(z^2+cz+c^2)$ gives the printed form up to branch constants.

The fluid vortex moves with the regular part of the complex velocity, excluding its own singular term. Since $w^{\prime }=u-iv$, its motion is the conjugated derivative. Differentiating the remaining terms and setting $z=c$ givesThe overbar on the time derivative is essential.

Write $c=r{e}^{i\theta }$. Using $c^3-{\bar{c}}^3=2i{r}^3\sin 3\theta$ and multiplying the velocity equation by $e^{i\theta }$ givesThus $\dot{r}=3\Gamma \cot 3\theta /(4\pi r)$ and $\dot{\theta }=-\Gamma /(4\pi r^2)$. Direct differentiation now yieldsHence $\boxed{r\sin (3\theta )=\text{constant}}$ along the trajectory while it stays inside the wedge.

Conformality and analyticity. Use the orientation-preserving convention for a conformal map, with a continuously differentiable, nonzero differential. Its Jacobian matrix must be a positive scale times a rotation,Thus $u_x=v_y$, $u_y=-v_x$, the Cauchy-Riemann equations; continuous partial derivatives then imply that $u+iv$ is a holomorphic function. If angle preservation is allowed to reverse orientation, the statement needs this qualification: $z\mapsto \bar{z}$ preserves unsigned angles but is an anticonformal map and is not a holomorphic function.

The mapping. Put $a=\pi /L$. The displayed components combine to giveIts derivative does not vanish in the open half-strip $x>0$, $0<y<L$, so it is a conformal map there. Also $\mathrm{Im}w=\sinh (ax)\sin (ay)>0$. The lower edge maps to $[1,\infty )$, the upper edge to $(-\infty ,-1]$, and the vertical edge to $[-1,1]$. The open half-strip maps bijectively onto the upper half-plane. Indeed, $\cosh \zeta =\cosh \eta$ implies $\zeta =\pm \eta +2\pi im$; the half-strip allows only equality. Surjectivity follows by solving $t+t^{-1}=2w$ and choosing the root $t=e^{az}$ with $|t|>1$ and $0<\arg t<\pi$. The two corners are boundary critical points, not interior failures of conformality.

Transformed boundary values. A conformal map preserves Laplace's equation up to a positive scale. Thus the transformed Dirichlet problem is to find a real harmonic function $U(u,v)$ on $v>0$ withThe conflicting values at the original corners are interpreted as one-sided edge limits.

Logarithmic solution. Choose the complex logarithms of $w\pm 1$ with arguments between zero and $\pi$ in the upper half-plane. A linear combination $F(w)=\alpha \log (w+1)+\beta \log (w-1)+\gamma$ is a complex-valued harmonic function, since it is a holomorphic function; the real physical potential must be a real or imaginary part. In particular, setThenOn the three real boundary intervals, the argument difference is respectively $0,\pi ,0$, which gives the required values. This is the bounded harmonic solution, with $U\to V_0$ at infinity; uniqueness holds in this bounded class by the maximum principle. Without a boundedness or far-field condition, the edge data alone do not give uniqueness: one may add $C\sinh (\pi x/L)\sin (\pi y/L)$.

Back substitution. Sincethe required continuous argument difference is $2\arctan [\sin (ay)/\sinh (ax)]$ in the open half-strip. ConsequentlyAs $y\to 0,L$ with $x>0$, the arctangent tends to zero; as $x\to 0^{+}$ with $0<y<L$, it tends to $\pi /2$. These limits verify every open edge of the Dirichlet problem.

A conformal map preserves the magnitude and orientation of angles between intersecting smooth curves locally. For a complex holomorphic function, the condition at $z_0$ is $f^{\prime }(z_0)\ne 0$: the first-order change $\delta w=f^{\prime }(z_0)\delta z$ rotates and uniformly scales every tangent vector by the same complex factor. This local statement does not require a globally one-to-one map.

A conformal map preserves local oriented angles and has a nonzero local scale factor. For an orientation-preserving complex map this is equivalent to holomorphy with $f^{\prime }(z)\ne 0$ at the point concerned. Infinitesimal figures are rotated and uniformly scaled there, but finite shapes need not be preserved: derivative magnitude and direction can vary across a region. Antiholomorphic maps preserve unoriented angles while reversing orientation; the convention being used must be stated.

For $z=x+iy$ in the strip,so $\mathrm{Im}f(z)>0$. This is a bijective conformal map onto the upper half-plane. One way to verify bijectivity is to factor it as $w=e^{\pi z/a}$, followed by $(w+w^{-1})/2$. The first map gives $\mathrm{Im}w>0$, $|w|>1$. The two inverse roots of the second map have product one; for any point in the upper half-plane exactly one root has modulus greater than one, and its imaginary part is positive because $\mathrm{Im}[(w+w^{-1})/2]=(1-|w{|}^{-2})\mathrm{Im}w/2$.

The three boundary pieces map as follows: the lower edge goes to $(1,\infty )$, the upper edge to $(-\infty ,-1)$, and the vertical edge to $(-1,1)$. Thus the desired exit event becomes exit from the upper half-plane to the left of $-1$. WriteBy conformal invariance of harmonic measure and the Cauchy exit law from a Brownian half-plane,The strip exit is finite almost surely, since its imaginary coordinate exits $(0,a)$ in finite time; corners have probability zero because individual points are polar points for planar Brownian motion. This is the Top-edge exit probability of a Brownian half-strip.

If $f:D\to D^{\prime }$ is a conformal map, the image of planar Brownian motion in $D$ is Brownian motion in $D^{\prime }$ after the increasing clock changeMore precisely, $f(Z_{A^{-1}(u)})$ is planar Brownian motion, starting at $f(Z_0)$ and run up to the relevant exit time. The Itô formula and the Cauchy-Riemann equations give image-coordinate local martingales with common bracket $A$ and zero cross bracket; the Lévy characterization of multidimensional Brownian motion identifies the time-changed process. In particular, conformal invariance of planar Brownian motion transports exit locations and harmonic measure, while changing traversal speed.

Let $\zeta =g(z)$ be a conformal map, and let a point vortex of circulation $\Gamma$ lie at $z=c$. Suppose the mapped complex potential of a two-dimensional potential flow iswhere $R$ is a holomorphic function near $g(c)$ and represents the other point vortices, boundaries, and any imposed flow. The physical complex velocity is $w^{\prime }=g^{\prime }{W}^{\prime }$. The point vortex follows its regular velocity, found by removing $\Gamma /[2\pi i(z-c)]$ before taking the limit. A Taylor expansion proveswhich gives the title formula. The complex conjugate occurs because the complex velocity convention is $w^{\prime }=u-iv$, whereas $\dot{c}=u+iv$.

For a mapped upper half-plane with no imposed flow, the method of images gives $R(\zeta )=-\Gamma \log (\zeta -\overline{g(c)})/(2\pi i)$, henceThe first term is essential: excluding a mapped singularity and then transforming its regular velocity does not in general equal subtracting the physical singularity after transformation.

For a point vortex of circulation $\Gamma$ in the wedge $0<\theta <\pi /n$, with positive integer $n$, the conformal map $z\mapsto z^n$ sends the domain to the upper half-plane. An opposite image vortex gives the complex potentialRemove the singular self-induced velocity before setting $z=c$. The resulting complex velocity isThe finite first term follows from the expansion of $(z^n-c^n)/(z-c)$ at $c$. In polar coordinates this becomes $\dot{r}=n\Gamma \cot (n\theta )/(4\pi r)$ and $\dot{\theta }=-\Gamma /(4\pi r^2)$. Therefore $r\sin (n\theta )$ is constant along a vortex trajectory. The case $n=1$ is motion parallel to a plane wall, while $n=3$ describes a sixty-degree wedge.

For $0<k<1$, principal upper-half-plane branches give a conformal map onto the upper half-plane cut along $[0,\tau ]$, where $\tau =2k^k(1-k{)}^{1-k}{e}^{i\pi k}$. Its derivative is $(z-1+2k)(z-1{)}^{k-1}(z+1{)}^{-k}$. The two prevertices $-1,1$ map to the slit base, and $1-2k$ maps to its tip. Boundary traversal goes out along the slit and back, while $g(z)=z+1-2k+O(1/z)$ at infinity. The argument principle gives one preimage of every point off the slit.

A prime end describes an approach to the boundary of a simply connected domain through a nested sequence of crosscuts. A conformal map induces a correspondence between prime ends, even when the Euclidean boundary does not have a continuous one-to-one parametrization. This is the natural boundary interpretation of the endpoints and growing tip of a Loewner chain.

With the square root positive for $Z$ on the positive imaginary axis, the displayed conformal map takes the upper half-plane bijectively to $\{\mathrm{Re}z>0\}\setminus [1,\infty )$. Squaring, subtracting from one, taking the positive-real-part square root, and reciprocating identifies each intermediate slit domain. The boundary prevertices $-1,0,1$ correspond to the lower infinite end, the slit tip at one, and the upper infinite end. The inverse is $\sqrt{1-z^{-2}}$ with positive imaginary part.

A conformal map from the complex upper half-plane to a polygonal domain with finite prevertices $a_j\in \mathbb{R}$ and interior angles $\pi {\alpha }_j$ has the displayed derivative, with upper-half-plane branches of the powers. Integrating supplies the map, up to translation and nonzero scaling. For a slit boundary, repeated geometric vertices are treated as distinct boundary approaches: the two sides at the slit base have their respective angles, and a slit tip has angle $2\pi$, hence exponent one.

The map $z\mapsto z^2$ is a bijective conformal map from the open first quadrant to the complex upper half-plane. In polar coordinates it doubles arguments from $(0,\pi /2)$ to $(0,\pi )$, and its derivative $2z$ never vanishes inside the quadrant. Its inverse is the branch of the square root with argument in $(0,\pi /2)$. By conformal invariance of harmonicity, a harmonic function $f$ on the half-plane pulls back to $g(x,y)=f(x^2-y^2,2xy)$. The two boundary rays map to the positive and negative real axes, respectively.

On $S_h=\{z:0<\mathrm{Im}z<h\}$, let $\beta \in S_h$ and write $e^{\pi \beta /h}=A+iB$ with $B>0$. The exponential function $E(z)=e^{\pi z/h}$ is a conformal map from the strip bijectively onto the upper half-plane, sending $ih/2$ to $i$. The affine map $w\mapsto A+Bw$ is an automorphism of the half-plane sending $i$ to $E(\beta )$. Conjugate it by $E$, using the principal complex logarithm as the inverse. The displayed formula is a conformal automorphism of the strip and satisfies $g_{\beta }(ih/2)=\beta$. The branch range $0<\mathrm{Im}\mathrm{Log}w<\pi$ is what prevents an unwanted vertical period in this inverse.

If a conformal map or anticonformal map $T$ is an involution preserving a curve family, its extremal length can be computed using invariant conformal metrics. Replace any density by ${\rho }_s=(\rho +T^{\ast }\rho )/2$. Every curve and its image have length at least $L_{\rho }$, hence $L_{{\rho }_s}\ge L_{\rho }$. Convexity of the square and change of variables give $A_{{\rho }_s}\le A_{\rho }$, proving the assertion. For reflection in the real axis, invariant means $\rho (z)=\rho (\bar{z})$.^[ref]

The upper half-plane is the open setIt contains precisely the complex numbers with positive imaginary part: $1+i$ is inside, $1$ lies on its boundary, and $1-i$ belongs to the lower half-plane. In the complex plane its boundary is the real line; in the Riemann sphere the boundary also includes infinity. It is a convex set, hence a simply connected domain, and the Cayley transform between the half-plane and disk makes it biholomorphic to the unit disc.

With the hyperbolic metric $ds=|dz|/\mathrm{Im}z$, the same domain becomes the Poincaré half-plane model. Its hyperbolic geodesics are vertical lines and upper semicircles whose centres lie on the real line. Under the Cayley transform, these become arcs of circles orthogonal to the unit circle, with diameters as the straight-line cases. ^[ref]

Matching colours identify each hyperbolic geodesic and its image. The vertical line through zero maps to a diameter; the vertical line through one maps to an arc ending at $-i$ and $1$. A semicircle from $-1$ to $1$ also maps to a diameter, whereas the semicircle from $0$ to $2$ maps to a curved arc. Their endpoints lie on the boundary, outside the domains: the endpoints are infinitely far from interior points in hyperbolic distance.

The upper half-plane is useful because its straight boundary and explicit upper half-plane automorphisms make conformal maps and hyperbolic geometry calculable. The picture explains why shortest hyperbolic geodesic routes can look curved in Euclidean coordinates; their lengths are measured with the position-dependent hyperbolic metric.

For historical context, in his letter to Klein of 22 September 1882, Poincaré constructs a quadrilateral bounded by four circles orthogonal to a fundamental circle, and studies transformations generating repeated copies. In modern language, the sides inside that circle are hyperbolic geodesics in a disk model. The letter directly exhibits the connection between transformation groups, geometry and complex functions. ^[ref]