Complex Analysis - Maths KU

74 Chapter 2 Complex Functions and Mappings
–8 –6 –4 –2
S1 8
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S
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–8 –6 –4 –2
S1 –2
S2 –4
8
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–8 –6 –4 –2
S 2
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S′
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(a) Rotation by π/2 (b) Magnification by 4 (c) Translation by 2 + 3i
Figure 2.14 Linear mapping of a rectangle
of mappings is depicted in Figure 2.14. In Figure 2.14(a), the rectangle S
shown in color is rotated through π/2 onto the rectangle S1 shown in black;
in Figure 2.14(b), the rectangle S1 shown in color is magnified by4 onto the
rectangle S2 shown in black; and finally, in Figure 2.14(c), the rectangle S2
shown in color is translated by2 + 3i onto the rectangle S ′ shown in black.
EXAMPLE 5 A Linear Mapping of a Triangle
Find a complex linear function that maps the equilateral triangle with vertices
1+i, 2+i, and 3
2 + � 1+ 1
√ �
2 3 i onto the equilateral triangle with vertices i,
√
3+2i, and 3i.
Solution Let S1 denote the triangle with vertices 1 + i, 2 + i, and 3
2 +
� √ �
1
′ 1+ 2 3 i shown in color in Figure 2.15(a), and let S represent the triangle
with vertices i, 3i, and √ 3+2i shown in black in Figure 2.15(d). There are
manyways to find a linear mapping that maps S1 onto S ′ . One approach is
the following: We first translate S1 to have one of its vertices at the origin.
If we decide that the vertex 1 + i should be mapped onto 0, then this is
accomplished bythe translation T1(z) =z − (1 + i). Let S2 be the image
of S1 under T1. Then S2 is the triangle with vertices 0, 1, and 1
√
1
2 + 2 3i
shown in black in Figure 2.15(a). From Figure 2.15(a), we see that the angle
between the imaginaryaxis and the edge of S2 √ containing the vertices 0 and
1 1
2 + 2 3i is π/6. Thus, a rotation through an angle of π/6 radians counterclockwise
about the origin will map S2 onto a triangle with two vertices on the
imaginaryaxis. This rotation is given byR(z) = � eiπ/6� z = � √
1 1
2 3+ 2i� z,
and the image of S2 under R is the triangle S3 with vertices at 0, 1
√
1
2 3+ 2i, and i shown in black in Figure 2.15(b). It is easyto verifythat each side of
the triangle S3 has length 1. Because each side of the desired triangle S ′ has
length 2, we next magnify S3 bya factor of 2. The magnification M(z) =2z
maps the triangle S3 shown in color in Figure 2.15(c) onto the triangle S4
with vertices 0, √ 3+i, and 2i shown in black in Figure 2.15(c). Finally, we
translate S4 by i using the mapping T2(z) =z + i. This translation maps the
4
2
–2
–4
2
4