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1 (1 t )(1 k 2t 2 )
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Now the transformation looks like Eq (105) To nd the vertices of the polygon, we look for the zeros of the integrand First consider 1 t 2 = 0 , which tells us that z = 1 Next we have 1 k 2t 2 = 0 from which it follows that z = (1 / k ) Since j / 1 each term in the Schwarz-Christoffel transformation is of the form ( z x j )
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Figure 104 The transformation in Example 101 maps a rectangle to the upper half plane
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and the exponents in this example are j / 1 = 1 / 2 , it follows that 1 = 2 = /2 , or each angle is increased by /2 at each vertex The polygon described by this is a rectangle The height of the rectangle is found from h=
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(1 t 2 ) 1/ 2 (1 k 2t 2 ) 1/ 2 dt
The width of the rectangle is W = 2 (1 t 2 ) 1/ 2 (1 k 2t 2 ) 1/ 2 dt
Looking at the de nition of the transformation, w(0) = 1 / (1 t 2 )(1 k 2 t 2 ) dt = 0, 0 so the origin of the w plane is mapped to the origin of the z plane The vertices of the polygon are located at ( W , 0),(W , 0),(W , ih), and ( W , ih) The transformation is illustrated in Fig 104
Summary
In this chapter, we wrapped up the discussion of mappings or transformations begun in Chap 9 First we stated the Riemann mapping theorem, which guarantees the existence of a mapping between a region of the w plane and a region of the z plane Next we introduced the Schwarz-Christoffel transformation, which allows us to map a polygon to the upper half plane
Quiz
dt 1 What type of region does the transformation w = A + B map to 1 t2 the upper half plane
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The Gamma and Zeta Functions
In this chapter, we review two important functions related to complex analysis, the gamma and zeta functions
The Gamma Function
The gamma function can be de ned in terms of complex variable z provided that Re(z) > 0 as follows: ( z ) = t z 1e t dt
(111)
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Complex Variables Demysti ed
We can show that Eq (111) is convergent in the right-hand plane by examining its behavior at t = 0 and as t Using | z | Re z we have t z 1e t t Re z 1 This tells us that Eq (111) is nite at t = 0 As t , note that we have t z 1e t t Re z 1e t e t / 2 This shows that the integral is convergent for large t If n is a positive integer, then (n + 1) = n(n 1) 1 = n! (112)
This follows from the recursion relation for the gamma function, which holds for any z (not just integers): ( z + 1) = z ( z ) (113)
EXAMPLE 111 Prove the recursion relation in Eq (113) for the gamma function SOLUTION This can be done using the de nition in Eq (111) We calculate ( z + 1): ( z + 1) = t z e t dt
Now notice what happens if we take the derivative of the integrand: d z t (t e ) = z t z 1e t t z e t dt It follows that t z e t = z t z 1e t d z t (t e ) dt
The Gamma and Zeta Functions
So the integral can be written as ( z + 1) = t z e t dt
0 d = z t z 1e t (t z e t ) dt 0 dt d = z t z 1e t dt (t z e t )dt 0 0 dt
Looking at the second term, we have
Therefore
d z t (t e )dt = (t z e t ) dt
t =0
= lim t z e t lim t z e t = 0 0 = 0
t t 0
( z + 1) = t z e t dt
= z t z 1e t dt
= z t z 1e t dt
( z + 1) = z ( z ) EXAMPLE 112 Show that an alternative de nition of the gamma function is given by ( z ) = 2 x 2 z 1e x dx
(114)
SOLUTION This is a simple substitution problem We start with Eq (111) and choose t = x2 It follows that dt = 2 xdx
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