1549312215-Complex_Analysis_for_Mathematics_and_Engineering_5th_edition__Mathews

54 CHAPTER 2 • COMPLEX FUNCTIONS

y !<z> = z^2 v

........ ........................ -........ -.. .. ....

.... .. ..

,..... .. .........

• --+----+ x --.+---!---• II

·. .......... .. .. , .........

...................................... ..

Figure 2.4 The function f(z) = z^2 is not one-to-one.

equation w = I (z) and "solve" for z as a function of w. Doing so produces an
inverse function z = g ( w) where the following equations hold:

g (! (z)) = z for all z E A, and

I (g(w)) = w for all w E B. (2-3)

Conversely, if w = f ( z) and z = g ( w) are functions that map A into B

and B into A, respectively, and Equations (2-3) hold, then I maps the set A

one-to-one and onto the set B.

Further, if I is a one-to-one mapping from D onto T and if A is a subset

of D , then I is a one-to-one mapping from A onto its image B. We can also

show that if ( = I (z) is a one-to-one mapping from A onto Band w = g (() is a

one-to-one mapping from B onto S, then the composite mapping w = g (! (z))

is a one-to-one mapping from A onto S.

We usually indicate the inverse of I by the symbol J-^1. If the domains of I

and 1-^1 are A and B, respectively, we can rewrite Equations (2-3) as

1-^1 (! (z)) = z for all z EA, and

I (f-^1 (w)) = w for all w EB.

Also, for zo EA and wo EB,

wo = I (zo) iff 1-^1 (wo) = zo.

(2-4)

(2-5)

• EXAMPLE 2.6 If w = I (z) = iz for any complex number z, find 1-^1 (w).
  Solution We can easily show I is one-to-one and onto the entire complex

plane. We solve for z, given w = I (z) = iz, to get z = T = -iw. By Equations

(2-5), this result implies that 1-^1 (w) = -iw for all complex numbers w.

Remark 2.2 Once we have specified 1-^1 (w) = -iw for all complex numbers

w, we note that there is nothing magical about the symbol w. We could just as

easily write 1-^1 (z) = -iz for all complex numbers z. •