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Fall 2009 - Problem 10

conformal mappings, construction

Consider the unit sphere in R3\R^3 with poles removed:

S≔{(cos⁡θsin⁡φ,sin⁡θsin⁡φ,cos⁡φ)∣0<φ<π and θ∈R}.S \coloneqq \set{\p{\cos\theta\sin\phi, \sin\theta\sin\phi, \cos\phi} \mid 0 < \phi < \pi \text{ and } \theta \in \R}.

Give an explicit formula for a conformal map from the complex plane onto SS so that horizontal lines are mapped to circles of constant φ\phi and vertical lines are mapped to arcs of constant θ\theta.
Solution.

Let f ⁣:C∖{0}→S\func{f}{\C \setminus \set{0}}{S} be given by stereographic projection, that is,

f(z)=(2Re⁡z∣z∣2+1,2Im⁡z∣z∣2+1,∣z∣2−1∣z∣2+1),f\p{z} = \p{\frac{2\Re{z}}{\abs{z}^2 + 1}, \frac{2\Im{z}}{\abs{z}^2 + 1}, \frac{\abs{z}^2 - 1}{\abs{z}^2 + 1}},

which is a conformal map sending circles centered at the origin to circles of constant φ\phi. Notice that ff has inverse

f−1(cos⁡θsin⁡φ,sin⁡θsin⁡φ,cos⁡φ)=cos⁡θsin⁡φ+i(sin⁡θsin⁡φ)1−cos⁡φ=(sin⁡φ1−cos⁡φ)eiθ\begin{aligned} f^{-1}\p{\cos\theta\sin\phi, \sin\theta\sin\phi, \cos\phi} &= \frac{\cos\theta\sin\phi + i\p{\sin\theta\sin\phi}}{1 - \cos\phi} \\ &= \p{\frac{\sin\phi}{1 - \cos\phi}}e^{i\theta} \end{aligned}

From here, we see that an arc of constant θ\theta comes from a straight line passing through the origin that makes an angle θ\theta with the real axis.

To find a map that sends any horizontal line to circles of constant φ\phi and vertical lines to arcs of constant θ\theta, it suffices to find a conformal map which:

  • sends horizontal lines to a circle centered at the origin and
  • sends vertical lines to a line passing through the origin.

The map exp⁡ ⁣:C→C∖{0}\func{\exp}{\C}{\C \setminus \set{0}} does the trick:

exp⁡(x+iy)=ex(cos⁡y+isin⁡y).\exp\p{x + iy} = e^x\p{\cos{y} + i\sin{y}}.

We see that horizontal lines, where xx is held constant, are mapped to circles of radius exe^x centered at the origin. Similarly, vertical lines, where yy is held constant, is mapped to a line which is contained in the line spanned by eiye^{iy}. Thus, f∘exp⁡ ⁣:C→S\func{f \circ \exp}{\C}{S},

(f∘exp⁡)(z)=(2Re⁡(ez)∣ez∣2+1,2Im⁡(ez)∣ez∣2+1,∣ez∣2−1∣ez∣2+1)\p{f \circ \exp}\p{z} = \p{\frac{2\Re\p{e^z}}{\abs{e^z}^2 + 1}, \frac{2\Im\p{e^z}}{\abs{e^z}^2 + 1}, \frac{\abs{e^z}^2 - 1}{\abs{e^z}^2 + 1}}

is a conformal map with the desired properties.