Complex Analysis

Math 312 Spring 1998  MWF 10:30am - 11:25am
Buckmire  Fowler 112

Class 17 (Friday February 20)

SUMMARY Linear transformations, inversion mappings, and bilinear transformations CURRENT READING Brown & Curchill, pages NEXT READING Brown & Curchill pages

Linear Transformation Let us re-consider the idea that functions of a complex variable w = f(z) represent a mapping from the complex z=(x,y) plane to the complex w = (u, v) plane Rotation: Rotation by an angle α w = f_{1}(z) = e^{iα}z Scaling: Magni�cation/reduction by a factor of |a| w = f_{2}(z) = az (a is a real number) Translation: Shifting by a vector of size (Re(B), Im(B)) w = f_{3}(z) = z + B What do all all these transformations have in common? What properties of the image get preserved?

\underline{Examples} Consider the circle C_{1} : |z − 1| = 1. Find a series of transformations which map C_{1} onto C_{2} : |z − \frac{3}{2}i| = 2 Sketch the series of transformations in the space below.

Using the relationship between composition of functions and successive transformations write down a single function which transforms C_{1} into C_{2}

So, any linear transformation can be written as a composition of , and The Inversion Mapping \frac{1}{z} The function w = \frac{1}{z} establishes a one-to-one correspondence between the nonzero points of the z and w planes. Remember |z|^{2} = z\overline{z}, so w = 1/z can be treated as two successive mappings

Z = \frac{1}{|z|^{2}}z, w = \overline{Z}

These mappings represent a  followed by a  .

\underline{Groupwork} Think about the points

a. (|z| > 1): exterior to the circle |z| = 1

b. (|z| < 1): interior to the circle |z| = 1

c. ON the circle |z| = 1

Where do each of these sets get mapped to in the w-plane using the �inversion transformation?� (To answer this question you should pick a point in each one of these sets and see where it is mapped under the w = 1/z transformation. You may want to split this job among the members of the group.)

What is the image of the point z = 0 under the inversion mapping? What is the image of the point at in�nity under the inversion mapping? To answer these questions you should recall how we deal with complex limits involving ∞

lim_{z→∞} f(z) = w_{0} ⟺ lim_{z→0}f (\frac{1}{z}) = w_{0}

PREGNANT PAUSE: Take 2 minutes and think about the concepts on this page. READ through the worksheet and formulate any questions you may have.

My Question is:

Properties of the Inversion Mapping The mapping w = 1/z maps the extended complex plane to itself on a one-to-one basis. The mapping 1/z transforms circles and lines into circles and lines

Lines passing thru the origin  ⟼  Lines passing thru the origin
Lines NOT passing thru the origin  ⟼  Circles passing thru the origin
Circles passing thru the origin  ⟼ Lines NOT passing thru the origin
Circles NOT passing thru the origin ⟼ Circles NOT passing thru the origin

Bilinear Transformations Consider transformations of the form

w = f(z) = \frac{az + b}{cz + d}

They are also known as bilinear transformations or M� obius transformations. It;s easy to see that you can re-write this to produce an expression of the form

Azw + Bz + Cw + D = 0

where A, B, C and D can be expressed in terms of a, b, c and d.

Bilinear Transformation as composite mapping Notice that if c = 0 then our bi-linear transformation (linear in z and w) becomes just a linear transformation in z. If c ≠ 0 then we can re-write w = f(z) as

w = \frac{a}{c} + \frac{bc − ad}{c} \frac{1}{cz + d}

To show this, all we have to do is remember polynomial division:

If we look at the linear fractional transformation this way, we can see that it can be written as a composition of two linear transformations and an inverse mapping.

w = cz + d,  w_{1} = \frac{1}{w},  w_{2} = \frac{a}{c} + \frac{bc − ad}{c}w_{1}

Find the composition of these three mappings above, so that w_{2} = T (z), and by so doing, show that T is a �LFT.�

Thus LFTs can be thought of as a followed by a followed by a .

Therefore, we know that LFT’s map circles and lines to and Properties of Linear Fractional Transformations

Let f be a M� obius transformation. Then

• f can be expressed as the composition of a �nite number of rotations, translations, magni�cations and inversions

• f maps the extended complex-plane to itself

• f maps the class of circles and lines to circles and lines

• f is conformal (i.e. f′(z) ≠ 0) at every point besides its pole

Poles and Fixed Points A pole (regular singularity) of a function is a point z_{0} where lim_{z→z_{0}} f(z) = ∞ A �xed point of a function f(z) is a point iz_{0} such that f(z_{0}) = z_{0}. That is the point gets mapped to the same spot in the w-plane.

Find the poles of T (z) = \frac{az + b}{cz + d}. How many poles does it have? How many �xed points does it have? (These answers should depend on a, b, c and d.)

If a line or circle passes thru the pole of T then it must be mapped to a shape that goes thru the point at in�nty. Why? What kind of shape would do that?

So, if a line or circle does NOT pass thru the pole of T it must get mapped to what kind of shape?

Where does T map the point at in�nity?

Inverses of LFTs Since T is a one-to-one mapping on the extended complex plane, it has an inverse. If you solve w = T (z) so that z = T^{−1}(w), then

T^{−1}(w) = \frac{−dw + b}{cw − a}, (ad − bc ≠ 0)

Note that T^{−1} is also an LFT. In general, if S and T are two LFTs, then S(T (z)) is also an LFT. \underline{Example} Find the image of the interior of the circle C : |z − 2| = 2 under the LFT given by w = f(z) = \frac{z}{2z − 8} Sketch the image and pre-image of C under w = f(z)