Example 12-17. (Representing function of a complex variable as mapping)
Consider the complex function
ω=f(z) = z^2 = (x+i y)^2 =x^2 + 2 ixy +i^2 y^2 = (x^2 −y^2 ) + i(2 xy ) = u+i v
The complex function ω=f(z) = z^2 defines the mapping
u=u(x, y ) = x^2 −y^2 and v=v(x, y ) = 2 xy
In polar coordinates with z=reiθ, the image point is ω=f(z) = z^2 =r^2 ei^2 θ=Reiφ so
that the mapping in polar form is given by
R=r^2 , and φ= 2θ
Knowing the transformation equations one can then construct special figures to give
various interpretations of this mapping. The figure 12-11 illustrates four selected
interpretations to illustrate the mapping ω=f(z) = z^2.
The first mapping illustrates two circles with radius r 1 and r 2 and their image
circles r^21 and r^22. The rays θ=αand θ=βmap to the image rays φ= 2αand φ= 2β.
The green region Smaps to the green region S′.
The second mapping illustrates the hyperpola x^2 −y^2 =u and 2 xy =v for the
selected values of u and v given by u 0 , u 1 and v 0 , v 1. These hyperbola map to
straight lines in the ω−plane.
The third mapping illustrates the line x= 0 mapping to the line segment A′B′,
where v= 0 , with u=−y^2 , B < y < A. The line y = 0 maps to the line segment
B′C′,where v= 0 and u=x^2 ,for 0 < x < x 1. The line x=x 1 ,C < y < D maps to the
parabola with parametric equations u=x^21 −y^2 , v = 2 x 1 y,C < y < D.
The fourth mapping illustrates the hyperbola x^2 −y^2 =u, for u= 0, 2 , 4 , 6 mapping
to the lines u= 0 , 2 , 4 , 6 and the hyperbola 2 xy =v, for v= 2 , 4 , 6 are mapped to the
lines v= 2, 4 , 6.
It is left as an exercise for you to make up additional figures to illustrate the
mapping ω=f(z) = z^2.
