1549312215-Complex_Analysis_for_Mathematics_and_Engineering_5th_edition__Mathews

2.5 • THE RECIPROCAL TRANSFORMATION W = ~ 89

y

• 1

w=-I

z

--

v

-2

Figure 2.24 The mapping w = ~ discussed in Example 2.23.

To study images of "generalized circles," we consider the equation

A (x^2 + y^2 ) +Bx+ Cy+ D = 0,

u

where A, B, C , and Dare real numbers. This equation represents either a circle

or a line, depending on whether A =/: 0 or A = 0 , respectively. Transforming the

equation to polar coordinates gives

Ar^2 + r (Bcos8 + Csin8) + D = O.

Using the polar coordinate form of the reciprocal transformation given in
Equation (2-31), we can express the image of the curve in the preceding equa-
tion as

A + p(Bcos- Csin) + Dp^2 = 0,

which represents either a circle or a line, depending on whether D =/: 0 or D = 0,

respectively. Therefore, we have shown that the reciprocal transformation w = ~
carries the class of lines and circles onto itself.

• EXAMPLE 2.24 Find the images of the vertical lines x = a and the hori-

zontal lines y = b under the mapping w = ~.

Solution Taking into account the point at infinity, we see that the image of

the line x = 0 is the line u = O; that is, the y-axis is mapped onto the v-axis.

Similarly, the x-axis is mapped onto the 1L-axis. Again, the inverse mapping is
z = ~ = .. •~v• + i ,,i-:i~.,.,, so if a =/: 0, the vertical line x = a is mapped onto
the set of (u, v) points satisfying .. •~v• =a. For (u, v) =/: (O, O), this outcome is
equivalent to

u2 - ~u+-1-+v 2 = (u-2-)2 + v2 = (2-)2,
a 4a2 2a 2a