1549312215-Complex_Analysis_for_Mathematics_and_Engineering_5th_edition__Mathews

1.6 • THE TOPOLOGY OF COMPLEX NUMBERS 39

y

+--------.x

Figure l.Zl The straight-line segment C joining zo to Zt.

Sol u t io n Refer to Figure 1.21. T he vector form of a line shows that the
direction of C is z 1 - zo. As zo is a point on C, its vector equation is

C : z (t) = zo + (z1 - zo) t, for 0 $ t $ 1, or (1-48)

C: z(t) = [xo + (x1 - xo)t] +i[Yo + (Y1 - yo)t], for 0 $ t $ 1.

Clearly one parametrization for - C is

-C:1(t)=z1+(zo- z1)t, for0:$t:$1.

Note that 'Y (t) = z (l - t), which illustrates a general principle: If C is a

curve parametrized by z (t) for 0 $ t ::; 1, then one parametrization for - C will

be7(t) = z(l-t), for 0::; t $ 1.

A curve C having the property that z (a)= z (b) is said to be a closed curve.
The line segment (1-48) is not a closed curve. T he range of z (t) = x (t) + iy (t),
where x(t) = sin2tcost, and y(t) = sin2tsint for 0 ::; t::; 2?T is a closed curve
because z(O) = (O, 0) = z (2?r). The range of z(t) is the four- leaved rose shown
in Figure 1. 22. Note that, as t goes from 0 to ~, the point is on leaf 1; from ~

y

Figure 1.ZZ The curvex(t) = sin2tcost, y(t) = sin2tsint for 0::; t::; 27T, which

forms a four-leaved rose.