1549312215-Complex_Analysis_for_Mathematics_and_Engineering_5th_edition__Mathews

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492 CHAPTER 11 • APPLICATIONS OF HARMONIC FUNCTIONS


will determine a mapping w = g(z ) of the upper half-plane onto the domain

indicated in Figure ll.74(a). With a 1 = ~. we let d -t oo, t hen a 2 --+ 7r and

a 3 -t - 2 " , and the limiting formula for the derivative g' ( z) becomes


J' (z) =Ai (z + 1 )-(w/2)/ w (z)-(w)/" (z-1)-(- w/ 2}/w


l(z- 1)^112 z- 1 z - 1
= A 1 - = A 1 =A ,

z (z + 1) 1/2 z (z2 - 1)1/ 2 z (1 - z2)1/ 2

where A = -iA1, which will determine a mapping w = f (z) from the upper
half-plane onto the channel as indicated in Figure ll.74(b). Using Table 11.2,
we obtain


f (z) = A[! dz -i J dz ]
(1-z2)! z (z^2 - 1)!

= A [ Arcsin z + iArcsin;] + B.

If we use the principal branch of the inverse sine function, then the boundary


values f (-1) = 0 and f (1) = 1 + i lead to the system

A (-- 2 'Tr - -i'Tr 2 ) + B = (^0) • and A ('Tr - 2 + -i'Tr) 2 + B = 1 + i •
which we can solve to obtain A = .! and B =
1



  • i. Hence the required solution
    'Tr 2
    is
    f()


lA. iA. 1 l+i

w = z = - rcsmz + - rcsm- + -

2

-.
'Tr 'Tr z


-------~EXERCISES FOR SECTION 11.9


  1. Let a and K be real constants with 0 < K < 2. Use t he Schwarz-Christoffel
    formula. to show that the function w = f (z) = (z -a)K maps the upper ha.lf-
    plane Im(z) > 0 onto t he sector 0 < arg 0 w < K1r shown in Figure 11.75.


Figure 11. 75
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