1.6 The Geometry of Complex Numbers

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The _rst thing to note is that as far as addition and scaling are concerned,

we are in R2, so there is nothing new. You can easily draw the line segment

t(2 􀀀 i3) + (1􀀀t)(7 + i4); t 2 [0; 1]

and if you do this in the point notation, you are just doing _rst year linear

algebra again. I shall assume that you can do this and don't _nd it very

exciting.

Life starts to get more interesting if we look at the geometry of multiplication.

For this, the matrix form is going to make our life simpler.

First, note that any complex number can be put in the form r(cos _+i sin _),

which is a real number multiplying a complex number of modulus 1. This

means that it is a multiple of some point lying on the unit circle, if we think

in terms of points in the plane. If we take r positive, then this expression is

unique up to multiples of 2_; if r is zero then it isn't. I shall NEVER take

r negative in this course, and it is better to have nothing to do with those

low-life who have been seen doing it after dark.

If we write this in matrix form, we get a much clearer picture of what is

happening: the complex number comes out as the matrix:

r _ cos _ 􀀀sin _

sin _ cos _ _

If you stop to think about what this matrix does, you can see that the r part

merely stretches everything by a factor of r. If r = 2 then distances from

the origin get doubled. Of course, if 0 < r < 1 then the stretch is actually a

compression, but I shall use the word 'stretch' in general.

The

_ cos _ 􀀀sin _

sin _ cos _ _

part of the complex number merely rotates by an angle of _ in the positive

(anti-clockwise) sense.

It follows that multiplying by a complex number is a mixture of a stretching

by the modulus of the number, and a rotation by the argument of the number.

1.6. THE GEOMETRY OF COMPLEX NUMBERS 27

3+4i

Figure 1.1: Extracting Roots

And this is all that happens, but it is enough to give us some quite pretty

results, as you will see.

Example 1.6.1 Find the _fth root of 3+i4

Solution The complex number can be drawn in the usual way as in _gure 1.1,

or written as the matrix

5 _ cos _ 􀀀sin _

sin _ cos _ _

where _ = arcsin 4=5. The simplest representation is probably in polars,

(5; arcsin 4=5), or if you prefer

5(cos _ + i sin _)

A _fth root can be extracted by _rst taking the _fth root of 5. This takes care

of the stretching. The rotation part or angular part is just one _fth of the

angle. There are actually _ve distinct solutions:

51=5(cos _ + i sin _)

for _ = _=5; (_+2_)=5; (_+4_)=5; (_+6_)=5; (_+8_)=5 , and _ = arcsin 4=5 =

arccos 3=5.

28 CHAPTER 1. FUNDAMENTALS

I have hopped into the polar and classical forms quite cheerfully. Practice

does it.

Exercise 1.6.1 Draw the _fth roots on the _gure (or a copy of it).

Example 1.6.2 Draw two straight lines at right angles to each other in the

complex plane. Now choose a complex number, z, not equal to zero, and

multiply every point on each line by z. I claim that the result has to be two

straight lines, still cutting at right angles.

Solution The smart way is to point out that a scaling of the points along

a straight line by a positive real number takes it to a straight line still, and

rotating a straight line leaves it as a straight line. So the lines are still lines

after the transformation. A rigid rotation won't change an angle, nor will a

uniform scaling. So the claim has to be correct. In fact multiplication by a

non-zero complex number, being just a uniform scaling and a rotation, must

leave any angle between lines unchanged, not just right angles.

The dumb way is to use algebra.

Let one line be the set of points

L = fw 2 C : w = w0 + tw1; 9t 2 Rg

for w0 and w1 some _xed complex numbers, and t 2 R. Then transforming

this set by multiplying everything in it by z gives

zL = fw 2 C : w = zw0 + tzw1; 9t 2 Rg

which is still a straight line (through zw0 in the direction of zw1).

If the other line is

L0 = fw 2 C : w = w00

+ tw01

; 9t 2 Rg

then the same applies to this line too.

If the lines L; L0 are at right angles, then the directions w1; w01

are at right

angles. If we take

w1 = u + iv and w01

= u0 + iv0

1.7. CONCLUSIONS 29

then this means that we must have

uu0 + vv0 = 0

We need to show that zw1 and zw01

are also at right angles. if z = x + iy,

then we need to show

uu0 + vv0 = 0 )(xu 􀀀 yv)(xu0 􀀀 yv0) + (xv + yu)(xv0 + yu0) = 0

The right hand side simpli_es to

(x2 + y2)(uu0 + vv0)

so it is true.

The above problem and the two solutions that go with it carry an important

moral. It is this: If you can see what is going on, you can solve some problems

instantly just by looking at them. And if you can't, then you just have to

plug away doing algebra, with a serious risk of making a slip and wasting

hours of your time as well as getting the wrong answer.

Seeing the patterns that make things happen the way they do is quite interesting,

and it is boring to just plug away at algebra. So it is worth a bit of

trouble trying to understand the stu_ as opposed to just memorising rules

for doing the sums.

If you can cheerfully hop to the matrix representation of complex numbers,

some things are blindingly obvious that are completely obscure if you just

learn the rules for multiplying complex numbers in the classical form. This is

generally true in Mathematics, if you have several di_erent ways of thinking

about something, then you can often _nd one which makes your problems

very easy. If you try to cling to the one true way, then you make a lot of

work for yourself.

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