2.2 The function w = z2

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We can get some idea of what the function w = z2 does by the same process.

I have put rather more dots in the before picture, _gure 2.4 and also made

it smaller so you could see the 'after' picture at the same scale.

The picture in _gure 2.5 shows what happens to the data points after we

square them. Note the greater concentration in the centre.

Exercise 2.2.1 Can you explain the greater concentration towards the origin?

Exercise 2.2.2 Can you work out where the sharp ends came from? Why

are there only two pointy bits? Why are they along the Y-axis? How pointy

are they? What is the angle between the opposite curves?

2.2. THE FUNCTION W = Z2 37

Figure 2.4: The square again

Figure 2.5: After Squaring the square

38 CHAPTER 2. EXAMPLES OF COMPLEX FUNCTIONS

Figure 2.6: A sector of the unit disk

Exercise 2.2.3 Try to get a clearer picture of what w = z2 does by calculating

some values. I suggest you look at the unit circle for a start, and see

what happens there. Then check out to see how the radial distance from the

origin (the modulus) of the points enters into the mapping.

It is possible to give you some help with the last exercise: in _gure 2.6 I have

shown some points placed in a sector of the unit disk, and in _gure 2.7 I

have shown what happens when each point is squared. You should be able

to calculate the squares for enough points on a calculator to see what is going

on.

Your calculations can sometimes be much simpli_ed by doing them in polars,

and your points should be chosen judiciously rather than randomly.

As an alternative, those of you who can program a computer can do what I

have done, and write a little program to do it for you. If you cannot program,

you should learn how to do so, preferably in C or PASCAL. MATLAB can

also do this sort of thing, I am told, but it seems to take longer to do easy

things like this. An engineer who can't program is an anomaly. It isn't

di_cult, and it's a useful skill.

Exercise 2.2.4 Can you see what would happen under the function w = z2

2.2. THE FUNCTION W = Z2 39

Figure 2.7: After Squaring the Sector

if we took a sector of the disk in the second quadrant instead of the _rst?

Exercise 2.2.5 Can you see what would happen to a sector in the _rst segment

which had a radius from zero up to 2 instead of up to 1? If it only went

up to 0.5?

Example 2.2.1 Can you see what happens to the X-axis under the same

function? The Y-axis? A coordinate grid of horizontal and vertical lines?

Solution

The program has been modi_ed a bit to draw the grid points as shown in

_gure 2.8. (If you are viewing this on the screen, the picture may be grottied

up a bit. It looks OK at high enough resolution). The squared grid points are

shown in _gure 2.9.

The rectangular grid gets transformed into a parabolic grid, and we can use

this for specifying coordinates just as well as a rectangular one. There are

some problems where this is a very smart move.

Note that the curves intersect at what looks suspiciously like a right angle. Is

it?

40 CHAPTER 2. EXAMPLES OF COMPLEX FUNCTIONS

Figure 2.8: The usual Coordinate Grid

Figure 2.9: The result of squaring all the grid points: A NEW coordinate

Grid

2.2. THE FUNCTION W = Z2 41

Exercise 2.2.6 Can you work out what would happen if we took instead the

function w = z3? For the case of a sector of the unit disk, or of a grid of

points?

It is rather important that you develop a feel for what simple functions do

to the complex plane or bits of it. You are going to need as much expertise

with Complex functions as you have with real functions, and so far we have

only looked at a few of them.

In working out what they do, you have a choice: either learn to program so

that you can do all the sums the easy way, or get very fast at doing them on a

calculator, or use a lot of intelligence and thought in deciding how to choose

some points that will tell you the most about the function. It is the last

method which is best; you can fail to get much enlightenment from looking

at a bunch of dots, but the process of _guring out what happens to lines and

curves is very informative.

Example 2.2.2 What is the image under the map f(z) = z2 of the strip of

width 1.0 and height 2.0 bounded by the X-axis and the Y-axis on two sides,

and having the origin in the lower left corner and the point 1 + i 2 at the

top right corner?

Solution

Let's _rst draw a picture of the strip so we have a grasp of the before situation.

I show this with dots in _gure 2.10. I have changed the scale so that the

answer will _t on the page.

Look at the bounding edges of our strip: there is a part of the X-axis between

0 and 1 for a start. Where does this go?

Well, the points are all of the form x + i0 for 0 _ x _ 1. If you square a

complex number with zero imaginary part, the result is real, and if you square

a number between 0 and 1, it stays between 0 and 1, although it moves closer

to 0. So this part of the edge stays in position, although it gets deformed

towards the origin.

Now look at the vertical line which is on the Y-axis. These are the points:

fiy : 0 _ y _ 2g

42 CHAPTER 2. EXAMPLES OF COMPLEX FUNCTIONS

Figure 2.10: A vertical strip

If you square iy you get 􀀀y2, and if 0 _ y _ 2 you get the part of the X

axis between 􀀀4 and 0. So the left and bottom edges of the strip have been

straightened out to both lie along the X-axis.

We now look at the opposite edge, the points:

f1 + iy : 0 _ y _ 2g

We have

(1 + iy)(1 + iy) = (1􀀀y2) + i(2y)

and if we write the result as u + iv we get that u = 1􀀀y2 and v = 2y. This

is a parametric representation of a curve: eliminating y = v=2 we get

u = 1􀀀

v2

4

which is a parabola. Well, at last we get a parabola in there somewhere!

We only get the bit of it which has u lying between 1 and -3, with v lying

between 0 and 4.

Draw the bits we have got so far!

2.2. THE FUNCTION W = Z2 43

Finally, what happens to the top edge of the strip? This is:

fx + i2 : 0 _ x _ 1g

which when squared gives

fu + iv : u = x2

􀀀 4; v = 4x; 0 _ x _ 1g

which is a part of the parabola

u =

v2

16 􀀀 4

with one end at 􀀀4 + i0 and the other at 􀀀3 + i4.

Check that it all joins up to give a region with three bounding curves, two of

them parabolic and one linear.

Note how points get 'sucked in' towards the origin, and explain it to yourself.

The points inside the strip go inside the region, and everything inside the

unit disk gets pulled in towards the origin, because the modulus of a square is

smaller than the modulus of a point, when the latter is less than 1. Everything

outside the unit disk gets shifted away from the origin for the same reason,

and everything on the unit circle stays on it.

The output of the program is shown in _gure 2.11 It should con_rm your

expectations based on a little thought.

Suppose I had asked what happens to the unit disk under the map f(z) = z2?

You should be able to see fairly quickly that it goes to the unit disk, but in

a rather peculiar way: far from being the identity map, the perimeter is

stretched out to twice its length and wrapped around the unit circle twice.

Some people _nd this hard to visualise, which gives them a lot of trouble;

fortunately you are engineers and good at visualising things.

Looking just at the unit circle to see where that goes: imagine a loop made

of chewing gum circling a can of beans.

If we take the loop, stretch it to twice its length and then put it back around

the can, circling it twice, then we have performed the squaring map on it.

44 CHAPTER 2. EXAMPLES OF COMPLEX FUNCTIONS

Figure 2.11: The Strip after Squaring

Before

After

Figure 2.12: Squaring the Unit Circle

2.2. THE FUNCTION W = Z2 45

This is shown rather crudely in the 'after' part of _gure 2.12. You have to

imagine that we look at it from above to get the loop around the unit circle.

Also, it should be smoother than my drawing. Don't shoot the artist, he's

doing his best.

If you tried to 'do' the squaring function on a circular carpet representing

the unit disk, you would have to _rst cut the carpet along the X-axis from

the origin to 1+i0. You need to take the top part of the cut, and push points

close to the origin even closer. Then nail the top half of the cut section to

the oor, and drag the rest of the carpet with you as you walk around the

boundary. The carpet needs to be made of something stretchy, like chewinggum

1. When you have got back to your starting point, join up the tear you

made and you have a double covering of every point under the carpet.

It is worth trying hard to visualise this, chewing-gum carpet and all.

Notice that there are two points which get sent to any point on the unit circle

by the squaring map, which is simply an angle doubling. The same sort of

thing is true for points inside and outside the disk: there are two points sent

to a + ib for any a; b. The only exception is 0, which has a unique square

root, itself.

This is telling you that any non-zero complex number has two square roots.

In particular, -1 has i and 􀀀i as square roots. You should be able to visualise

the squaring function taking a carpet made of chewing-gum and sending two

points to every point.

This isn't exactly a formal proof of the claim that every non-zero complex

number has precisely two distinct square roots; there is one, and it is long

and subtle, because formalising our intuitions about carpets made of chewinggum

is quite tricky. This is done honestly in Topology courses. But the idea

of the proof is as outlined.

I have tried to sketch the resulting surface just before it gets nailed down. It

is impossible to draw it without it intersecting itself, which is an unfortunate

property of R3 rather than any intrinsic feature of the surface itself. It is

most easily thought of as follows; take two disks and glue them together at

the centres. In _gure 2.13, my disks have turned into cones touching at the

vertices. Cut each disk from the centre to a single point on the perimeter

in a straight line. This is the cut OP and OP' on the top disk, and the cut

1You need a quite horrid imagination to be good at maths.

46 CHAPTER 2. EXAMPLES OF COMPLEX FUNCTIONS

O

P P’

Q Q’

Figure 2.13: Squaring the Unit Disk

OQ, OQ' on the lower disk. Now join up the cuts, but instead of joining the

bits on the same disks, join the opposite edges on opposite disks. So glue

OP to OQ' and OP' to OQ. The fact that you cannot make it without it

intersecting itself is because you are a poor, inadequate three dimensional

being. If you were four dimensional, you could do it. See:

http://maths.uwa.edu.au/~mike/PURE/

and go to the fun pages. If you don't know what this means, you have never

done any net sur_ng, and you need to.

This surface ought to extend to in_nity radially; rather than being made

from two disks, it should be made from two copies of the complex plane

itself, with the gluings as described. It is known as a Riemann Surface.

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