\Ve rcprcsem 1he unil vec1or which is rnngem 10 a smoolh

Chapter 0, Problem 11.1

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\Ve rcprcsem 1he unil vec1or which is rnngem 10 a smoolh arc C at a poim : 0 by 1he complex number t. and we lei the number r dcno1e lhc unil vcclor tan gem 10 the image r of Cal lhc corresponding point u1 0 under a 1ransfonnation 11: =/(:).\Ve assume lluH f is analy1ic al : 0 and lhal /'(: 0 ) :f: 0. According lo Sec. 112. ( I ) argr = arg_((:o) + argt. In particular. if C is a scgmclll of the x axis with positive sense lO the right. thcn t = and arg t = 0 al each point : 0 = x on C. In lhal case. equation (I) becomes (2) arg r = arg _((.r ). Ir f' (:) has a constant argumem along that segment. it follows lhal arg r is constant. Hence the image r of C is also a segment of a straight line. 393 394 TI IE SCI IWARZ-CI IRISTOl'l:EL TR.\l\SIDRMA TIO!'\ CllAP. 11 Let us now construct a transformation ui = .f (:)that maps the whole .r axis onto a polygon of 11 sides. whcrc.r,. x2 . X11- 1. and oo arc the points on that axis whose images arc to be the vertices of the polygon and where XI < .\'2 < .. ' < X,, -I The vertices arc the /1 points U'i = f(xi) (j = 1. 2 ..... 11 - I) and 11 1 11 = .f(oo). The function f should be such that arg f' (:)jumps from one constant value to another at the points: = x j as the point:. traces out the x axis (Fig. I 80). \' I FIGURE 180 . r .\,, I . l' "'1 II If the function I is chosen such that .f'(:)=J\(:-.r1) /.;!(:-.r2) k:(:-.r11. J) /.:.. .. '. where J\ is a complex constant and each k i is a real constant. then the argument of f' (:) changes in the prescribed manner as : describes the real axis. This is seen by writing the argument of the derivative n) as ( 4) arg /' (: ) = arg J\ - k 1 arg (: - .r 1 ) - k2 arg(: - .r 2 ) - -k,, . 1 arg(: - x,, . 1). \Vhcn: = x andx < .r1. a ro ( - r ) - uo ( - r ) - - 1 rr> ( - r ' e ... - . t - ' e .... - . 2: - . - ' e ..... - . " J) = :r. When .r 1 < x < .r2 the argument mg(: - .r 1) is 0 and each of the other arguments is ;r. According to equation (4). then. arg .('(:)increases ahruptJy hy the angle k1;r as: moves lo the right through the point: = .r 1. It again jumps in value. hy the amount k 2:r. as: passes through the point .r2, etc. In view of equation (2). the unit vector r is constant in direction as: moves from xi 1 to .r;: the point w thus moves in that fixed direction along a straight line. The direction of r changes ahmptly. hy the angle k;n. at the image point w; of x;. as shown in Fig. 180. Those angles k;n arc the extc1ior angles of the polygon described by the point u. The exterior angles can be limited to angles between -;r and :r. in which case - I < k 1 < I. Vole assume that the sides of the polygon never cross one another and that the polygon is given a positive. or counterclockwise. orientation. The sum of the cxtc1ior angles of a dosed polygon is. then. 2-:r: and the exterior angle at the vertex w11 SEC. 128 SCll\VARZ-CI IRISTOFFEI. TRAl'\Sl-ORM..\TIO:\ 395 '''hich is the image of the point:: = oo. can be wrillen Thus the numbers k i must necessarily satisfy the conditions (5) k I -f- k1 + -1- k11 -I -1- kn = 2. - I < ki < I U=l.2 ..... 11). Note that k" = 0 if (6) k I + k :!. + + k11 _ J = 2. This means that the direction of r docs not change at the point 11 1 11 So 11 1 11 is not a vertex. and the polygon has /1 - I sides. The existence of a mapping function .f whose derivative is given by equation (3) \viii be established in the next section.