Complex Analysis - Maths KU

70 Chapter 2 Complex Functions and Mappings
θ
φ
R(z)
Figure 2.10 Rotation
z
π/4
C′
Figure 2.11 Image of a line under
rotation
r
R
C
Figure 2.10. Observe also from (3) that an argument of R(z) isθ + φ, which
is θ radians greater than an argument of z. Therefore, the linear mapping
R(z) =az can be visualized in a single copyof the complex plane as the
process of rotating the point z counterclockwise through an angle of θ radians
about the origin to the point R(z). See Figure 2.10. Clearly, this increases
the argument of z by θ radians but does not change its modulus. In a similar
manner, if Arg(a) < 0, then the linear mapping R(z) =az can be visualized in
a single copyof the complex plane as the process of rotating points clockwise
through an angle of θ radians about the origin. For this reason the angle
θ =Arg(a) is called an angle of rotation of R.
EXAMPLE 2 Image of a Line under Rotation
Find the image of the real axis y = 0 under the linear mapping
R(z) =
� 1
2
√ √ �
1 2+ 2 2 i z.
Solution Let C denote the real axis y = 0 and let C ′ denote the image of C
under R. Since � √ √ �
� 1 1
2 2+ 2 2 i� = 1, the complex mapping R(z) is a rotation.
In√ order √to determine the angle of rotation, √ √we write the complex number
1 1
1 1
iπ/4
2 2+ 2 2 i in exponential form 2 2+ 2 2 i = e . If z and R(z) are
plotted in the same copyof the complex plane, then the point z is rotated
counterclockwise through π/4 radians about the origin to the point R(z). The
image C ′ is, therefore, the line v = u, which contains the origin and makes an
angle of π/4 radians with the real axis. This mapping is depicted in a single
copyof the complex plane in Figure 2.11 where the real axis shown in color
is mapped onto the line shown in black by R(z) = � √ √ �
1 1
2 2+ 2 2i z.
As with translations, rotations will not change the shape or size of a figure
in the complex plane. Thus, the image of a line, circle, or triangle under a
rotation will also be a line, circle, or triangle, respectively.
Magnifications The final type of special linear function we consider
is magnification. A complex linear function
M(z) =az, a > 0, (4)
is called a magnification. Recall from the Remarks at the end of Section 1.1
that since there is no concept of order in the complex number system, it is
implicit in the inequality a>0 that the symbol a represents a real number.
Therefore, if z = x + iy, then M(z) =az = ax + iay, and so the image of the
point (x, y) is the point (ax, ay). Using the exponential form z = re iθ of z,
we can also express the function in (4) as:
M(z) =a � re iθ� =(ar) e iθ . (5)