1549312215-Complex_Analysis_for_Mathematics_and_Engineering_5th_edition__Mathews

10.1 • BASIC PROPERTIES OF CONFORMAL MAPPINGS 399

y

IV :f(z)

v

T* l

Figure 10.3 The analytic mapping w
0,... , J (k - I ) (zo) = 0 and J(k) (zo) =f 0.

f(z) a t point zo, where f'(zo)

• EX AMPLE 10.2 Show that the mapping w = f (z) = z^2 maps the unit

square S = { x + iy : 0 < x < 1, 0 < y < 1} onto the region in the upper half-

plane Im ( w) > 0, which lies under the parabolas
'
1 1 2

u = 1 - -v^2 and u = - 1 + - v ,

4 4

as shown in Figure 10.4.

Solution The derivative is f' (z) = 2z, and we conclude that the mapping
w = z^2 is conformal for all z =f 0. Note that the right angles at the vertices

z1 = 1, z 2 = 1 + i, and z3 = i are mapped onto right angles at the vertices

wi = 1, w2 = 2i, and w3 = -1, respectively. At the point zo = 0, we have

f' (0) = 0 and f" (0) =f 0. Hence angles at the vertex zo = 0 are magnified by
the factor k = 2. In particular, the right angle at zo = 0 is mapped onto the
straight angle at wo = O.

Another property of a conformal mapping w = f (z) is obtained by consid-

ering the modulus off' (zo). If z 1 is near z 0 , we can use Equation (10-1) and

neglect the term fJ (z 1 ) (z 1 - zo). We then have the approximation

W1 - Wo = f (zi) - I (zo) "'"f I (zo) (z1 -zo). ( 10 -9)