1549312215-Complex_Analysis_for_Mathematics_and_Engineering_5th_edition__Mathews

410 CHAPTER 10 • CONFORMAL MAPPING

1 2. If S1 (z) = ::;~ and S2 (z) = .~ 3 , find S1 (82 (z)) and 82 (81 (z)).

13. Find the image of the quadrant x > 0, y > 0 under w = ::;~.

14. Show t hat Equation (10- 1 8) ca.n be written in the form of Equation (10-13).

1 5. Find the image of the horiwntal strip 0 < y < 2 under w = .:.,.

16. Show that the bilinear transformation w = S (z) = ~:t! is conformal at all points

z =I -.d.

17. A fixed poi11t of a mapping w = f (z) is a point zo such t hat f (zo) = ZQ. Show

t hat a bilinear transformation can have at most two fixed points.

1 8. Find the fixed points of

( ) a w = z:z - 1 +i ·

(b) w =^4 2z•±3- l.

10.3 Mappings Involving Elementary Functions

In Section 5 .1 we showed that the function w = f (z) = expz is a one-to-one
mapping of the fundamental period strip - 71' < y :=:; 11' in the z plane onto
thew plane with the point w = 0 deleted. Because f' (z) =f 0, the mapping

w = exp z is a conformal mapping at each point z in the complex plane. The

family of horizontal lines y = c for -'11' < c :=:; 11' and the segments x = a for

-11' < y :=:; 11' form an orthogonal grid in the fundamental period strip. Their

images under the mapping w = exp z are the rays p > 0 and </> = c and the

circles lwl = e", respectively. These images form an orthogonal curvilinear

grid in the w plane, as shown in Figure 10.9. If -71' < c < d :=:; '11', then the

rectangle R = { x + iy : a < x < b, c < y < d} is mapped one-to-one and onto

the region G = {peioi>: e" < p < eb, c < </> < d}. The inverse mapping is the

principal branch of the logarithm z = Logw.

Io this section we show how compositions of conformal transformations are

used to construct mappings with specified characteristics.

• EXAMPLE 10.8 Show that the transformation w = f (z) = :::;:; is a one-
  to-one conformal mapping of the horizontal strip 0 < y < 11' onto the disk lwl < l.
  Furthermore, the x-axis is mapped onto the lower semicircle bounding the disk,
  and the line y = 11' is mapped onto the upper semicircle.

Solution The function f is the composition of Z = expz followed by w = ~:;;:.

The transformation Z = exp z maps the horizontal strip 0 < y < 11' onto the

upper half-plane Im ( Z) > O; the x-axis is mapped on to the positive X-axis;

and the line y = 11' is mapped onto the negative X-axis. Then the bilinear

transformation w = ~:;:~ maps the upper half-plane Im (Z) > 0 onto the disk