396 CHAPTER 10 • CONFORMAL MAPPING
Let C: z (t) = x (t) + iy (t), - 1 :St :S 1 d enote a smooth curve that passes
through the point z(O) = z 0. A vector T tangent to Cat the point zo is given
by
T = z' (O),
where the complex number z' (0) is expressed as a vector.
The angle of inclination of T with respect to the positive x-axis is
/3 = Argz' (0).
The image of C under the mapping w = f (z) is the curve K given by the formula
K: w(t) = u(x(t),y(t)) +iv(x(t),y(t)). We can use the chain rule to show
that a vector T• tangent to Kat the point wo = f (zo) is given by
T " = w' (0) = f' (zo) z' (0).
The angle of inclination of T• with respect to the positive u-axis is
'Y = Arg f' (zo) + Argz' (0) = a+ /3, (10-2)
where a = Arg f' ( z 0 ). Therefore, the effect of the transformation w = f ( z) is
to rotate the angle of inclination of the tangent vector T at zo through the angle
a = Arg f' (zo) to obtain the angle of inclinat ion of the tangent vector T * at
wo. This situation is illustrated in Figure 10.L
A mapping w = f (z) is said to be angle preserving, or conformal at zo, if it
preserves angles between oriented curves in magnitude as well as in orientation.
T heorem 10 .1 shows where a mapping by an analytic function is conformal
w =f(z)
~
v
-o-----------11
Figure 10.1 The tangents at the points Zo and wo, where J is an analytic function and
f I (zo) =/: 0.