Solution 5.

See text and/or instructor's solution manual.

Solution.   The transformation    maps the point      onto the point   .

For the function     we have   .   Since   ,

by Theorem 10.1 in Section 10.1 , the mapping is conformal at    and "angles are preserved."

The ray    at    make an angle  .   In Section 10.2, we proved that a bilinear transformation maps the class of half-planes and disks onto itself.

Hence, the image of a line through the origin under the Möbius transformation      is a "circle."    Since   ,   and

the image of the ray   will be an arc     of a circle that passes through the points  .

Therefore, the arc      is inclined at the angle    at the point   .

We are done.

Aside.  The image of the point will be the point      on the arc     as shown by the calculation

We are really done.

Aside.  We can let Mathematica double check our work.

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We are really really done.

Aside.  We can explore some graphs.

The image of the ray    under      is an arc    of a circle that passes through    and  .

If the ray   make an angle    at    then the arc    is inclined at the angle    at the point  .

This solution is complements of the authors.

(c) 2008 John H. Mathews, Russell W. Howell