Conformal Mappings 203corresponding to some level sets of u and v. Show that, at every point of
intersection, the angle between Cu and Cv is TT/2. Hint: use (4.2.2) on page
192.
In the complex plane, the vector representation of a smooth curve is
equivalent to denning a continuously differentiable complex-valued function
z = z(t) of a real variable t, so that z(t) ^ 0; the tangent vector to the
curve is represented by the function z(t). Using the properties of complex
numbers, we conclude that if z\ — z(€) and z 2 = z 2 (s) represent two
smooth curves and zi(to) = z 2 (s 0 ), then, according to Exercise 4.1.4 on
page 184, the angle 8 between the curves is
6 = min (|Arg(ii(t 0 )) - Arg(i 2 (s 0 ))|,7r - |Arg(ii(t 0 )) - Arg(i 2 (s 0 ))|).
(4.2.18)
We will now see that an analytic function / — f(z) with non-zero deriva-
tive does not change the angle between two curves. Indeed, let z\ — z(t)
and z 2 = Z2(s) represent two smooth curves. The function / maps these
curves to wi(t) = f{z(t)) and w 2 (s) = f(z 2 (s)). By the chain rule,
Mt) = f'(zi(t))zi(t), w 2 (s) = f'(z 2 (s))z 2 (s). If z = Zl(t 0 ) = z 2 (s 0 )
is the point of intersection of the original curves, then f(z) is the point of
intersection of the images of these curves under /. By Exercise 4.1.4, the
argument of the product of two non-zero complex numbers is equal to the
sum of the arguments, so that, under the mapping /, all smooth curves that
pass through the point z are mapped onto curves w(t) = f(z(t)), which
are turned by the same angle Arg(/'(z)). Then relation (4.2.18) implies
that the angle between the original curves at the point z is the same as the
angle between the images at the point f(z). Note that the above calcula-
tions do not go through if f'(z) — 0, because in that case the argument of
f'(z) is not defined.
EXERCISE 4.2.23? The derivative of the analytic function f(z) = z^2 is equal
to zero when z = 0. By considering two lines passing through the origin,
show that this function doubles angles between curves at z = 0. Hint: a line
through the origin is defined by Arg(z) = const, and Arg(z^2 ) = 2Arg(z).
Recall that, to every mapping from R^2 to E^2 , we associate the Jacobian,
the function describing how the areas change locally at every point (see page
148 for a three-dimensional version). If written f{z) = u(x,y) + iv(x,y),
every function of a complex variable defines a mapping from E^2 to E^2 by
sending a point (x,y) to the point (u(x,y),v(x,y)).