Conformal Maps and Mobius Transformations

Let f be holomorphic near a. A tangent vector v at a maps to f'(a)v. If f'(a) is nonzero, this multiplication rotates every tangent by the same angle and scales every length by the same positive factor. It preserves oriented angles between regular curves. This is local conformality. The exponential qualifies at every finite point: its derivative e^z never vanishes.

For f(z)=z^2, the derivative 2z vanishes only at 0. Squaring sends a ray of direction theta to direction 2*theta, so it does not preserve angles at 0. Local conformality also differs from global injectivity. The exponential is conformal everywhere, but e^(z+2*pi*i)=e^z. In particular, 0 and 2*pi*i both map to 1.

A Mobius map is T(z)=(az+b)/(cz+d), where ad-bc != 0. Its derivative is T'(z)=(ad-bc)/(cz+d)^2, nonzero wherever cz+d != 0. A zero determinant instead gives a constant ratio where defined, provided the denominator is not identically zero. Multiplying all four coefficients by one nonzero number leaves the map unchanged.

The extended complex plane adds one point, infinity. A Mobius map is a bijection of this plane. If c != 0, its pole -d/c maps to infinity and T(infinity)=a/c, found by a limit. If c=0, infinity maps to infinity and T is affine. Mobius maps send generalized circles to generalized circles: these are circles and lines completed by infinity.

Two ordered triples of distinct extended-plane points determine exactly one Mobius map matching corresponding entries. For finite p,q,r, A(z)=((z-p)/(z-r))*((q-r)/(q-p)) sends p,q,r to 0,1,infinity in that order. Normalize the target triple with B in the corresponding order. Then T=B^-1 composed with A gives the prescribed images. Use limits when infinity occurs. Distinct targets are necessary because the map is injective.

Evaluate a ratio by substituting into both numerator and denominator. For R(z)=(z-2)/(z+2), R(4)=2/6=1/3. For a complex denominator, multiply both parts by its conjugate. With C(z)=(z-i)/(z+i) and real x, this gives C(x)=(x-i)^2/(x^2+1)=(x^2-1-2ix)/(x^2+1).

C(z)=(z-i)/(z+i) maps H={Im z>0} bijectively onto D={|w|<1}, with C(i)=0. For z=x+iy, |C(z)|^2=[x^2+(y-1)^2]/[x^2+(y+1)^2]<1 when y>0. Solving w=C(z) gives C^-1(w)=i(1+w)/(1-w). Its imaginary part is (1-|w|^2)/|1-w|^2>0 on D. Every disc point therefore has a unique inverse in H.

A real-valued harmonic u satisfies Delta u=u_xx+u_yy=0. A Dirichlet problem prescribes its boundary values. For holomorphic C, Delta(u composed with C)=|C'|^2*(Delta u) composed with C. For C:H to D, set disc data h=g composed with C^-1. Once a harmonic disc solution u with data h is known, v=u composed with C solves the half-plane problem. Apply the forward map C before u.

For C(z)=(z-i)/(z+i), finite real x have |C(x)|=1 and C(x) != 1. Their images form the unit circle minus 1; C(infinity)=1 fills the gap. Continuous data on the completed real boundary must have a common limit g(infinity) as x tends to either real infinity. The transformed data then satisfy h(1)=g(infinity). A disc solution continuous on the closed disc pulls back to these boundary values.