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774 CHAPTER 23 Conformal Mappings and Applications

passes over x 2 ,arg( f (z)) increases by α 2 π, then remains at this value until z passes over x 3 ,at

which arg( f (z)) increases by α 3 π. In general, arg( f (z)) increases by α j π as z passes over x j .

The net result (as z traverses the entire line) is that the real axis is mapped to a polygon P having
∗
exterior angles α 1 π,α 2 π,··· ,α n π, which are the same as P. In fact α 1 ,··· ,α n−1 determine all n
of these exterior angles, since
n

α j π = 2π.
j=1
∗
Now P has the same exterior angles as P but need not be the same as P because of its location
and size. We may have to rotate and/or magnify P to obtain P. These effects are achieved by
∗
∗
choosing x 1 ,··· , x n to make P similar to P, then choosing a (a rotation/magnification) and b (a
∗
translation) to superimpose P onto P.
If we choose x n =∞, then x 1 , x 2 ,··· , x n−1 ,∞ are mapped to the vertices of P. In this case,
the Schwarz-Christoffel transformation has the possibly simpler form as
z
f (z) = a (ξ − x 1 ) −α 1 (ξ − x 2 ) −α 2 ···(ξ − x n−1 ) −α n−1 dξ + b (23.4)
z 0
It can be shown that any conformal mapping of U onto a region bounded by a polygon must
have the form of a Schwarz-Christoffel mapping. In practice, it may be impossible to carry out
the integration needed to write a Schwarz-Christoffel transformation in closed form.

EXAMPLE 23.17
Suppose we want to map the upper half-plane onto a rectangle. Choose x 1 = 0, x 2 = 1,
and x 3 as any number greater than 1. The corresponding Schwarz-Christoffel transformation
equation (23.4) has the form
z 1
f (z) = a √ dξ + b
ξ(ξ − 1)(ξ − x 3 )
z 0
with a and b chosen to fit the dimensions and orientation of the original rectangle. The radical
appears because the exterior angles of a rectangle are all π/2, so each α j = 1/2. This integral is
an elliptic integral and cannot be evaluated in closed form.

EXAMPLE 23.18
We will map U onto the strip S defined by Im(w) > 0,−c < Re(z)< c in the w-plane. Here c is
a positive constant.
U and S are shown in Figure 23.30. To use the Schwarz-Christoffel transformation, think of
S as a polygon with vertices −c, c, and ∞. Choose x 1 =−1tomap −c to −1 and x 2 =1tomapto
c.Map ∞ to ∞. The exterior angles of S are π/2 and π/2, so α 1 = α 2 = 1/2. The transformation
has the form
z

w = f (z) = a (ξ + 1) −1/2 (ξ − 1) −1/2 dξ + b.
z 0
Choose z 0 = 0 and b = 0, and write
(ξ − 1) −1/2 =[−(1 − ξ)] −1/2 =−i(1 − ξ) −1/2 .

Writing −ai = A,wehave
z 1
w = f (z) = A dξ.
0 (1 − ξ )
2 1/2

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October 14, 2010 15:39 THM/NEIL Page-774 27410_23_ch23_p751-788