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Unformatted text preview: (w - w3)(w2 - w1) (z - z3)(z2 - z1)
Since plugging in the values (zi, wi) make it hold, it must be the one we're after.
(A) Mapping the upper half-plane onto the unit disc.
Since we need only say what happens to three points, we shall choose them to be z = -1,
0 and 1 on the real axis. Since these points are on the boundary of the half-plane, they
must be mapped to points on the boundary of unit disc, and we let -1Æ-1, 0Æ-i, 1Æ1
under the mapping. Substituting in (*) and solving for w gives:
-iz + 1
Question Why does it do what we claimed?
24 Answer We use the fact that lines or circles go to lines or circles.
1. Because of what happens to the three points, we know that the real axis Æ unit circle.
2. By continuity, nearby parallel lines must also go to (nearby) circles.
3. We know 0Æ-i. Also, we can check that iÆ0. Thus the positive imaginary axis goes
to the line starting at -i and going up.
4. By 2 & 3, lines above the real axis must go to circles inside the given circle.
5. Since Ï goes to i,(look at the highest powers of z) all these circles must touch the...
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