1.2. The Complex Plane
Complex numbers are defined as tuples (x,y) or equivalently x+iy. Therefore is is natural to visualize them as objects in the two-dimensional plane.
Definition 1.2.1: The Complex Plane The field of complex numbers is represented as points or vectors in the two-dimensional plane. If z = (x,y) = x+iy is a complex number, then x is represented on the horizonal, y on the vertical axis. The horizontal axis is called real axis while the vertical axis is the imaginary axis.
The complex plane
Visually, C looks like R2, and complex numbers are represented as "simple" 2-dimensional vectors. Even addition is defined just as addition in R2. The big difference between C and R2, though, is the definition of multiplication. In R2 no multiplication of vectors is defined. To be precise, there is the idea of dot-product and (if you think of R2 as embedded in R3) cross-product, but the definition of complex multiplication gives C its special structure.
Before we explore multiplication further, let's define a few easier operations in C and see their impact in the complex plane.
Example 1.2.2: Multiplication in 2-D and C Explore the differences between complex multiplication in C, the dot-product in R2, and the cross-product in R2 as embedded in R3.
Definition 1.2.3: Re, Im, Conj, and Modulus If z = x + iy is a complex number, then we define the following functions:
- The Real Part
- Re: C R, Re(z) = Re(x+iy) = x is the real part of z
- The Imaginary Part
- Im: C R, Im(z) = Im(x+iy) = y is the imaginary part of z
- The Conjugate or z-bar
- conj: C C, conj(z) = conj(x+iy) = x - iy is the conjugate of z. The conjugate of z is sometimes also denoted as (z-bar)
- The Modulus, or Absolute Value
- |.|: C R, |z| = |x+iy| = is the modulus, absolute value, or length of z
Geometrically, these operations are easy to visualize:
Here are a few concrete (simple) examples:
Example 1.2.4: Complex numbers in the plane Let u = 3 - 4i and v = i, and w = -2 Find algebraically and visually:
Our new operations can also be composed with one another:
Example 1.2.5: Complex operations composed Show that
- Re(Re(z)) = Re(z)
- Re(Im(z)) = Im(z)
- Im(Re(z)) = Im(Im(z)) = 0
Similar relations hold when composing the conjugate function, but since we will need these identities frequently we will state them in the form of a proposition:
Proposition 1.2.6: The conjugate operator For every complex number z we have:(1) |z| = |conj(z)| = ||
(2) |z|2 = z*conj(z) = z
(3) conj(conj(z)) = = z
Addition and in particular multiplication are what really define the structure of the complex plane and we need to figure out how to visualize those operations. Addition (and subtraction) will be familiar from vectors in the plane:
Proposition 1.2.7: Adding complex numbers geometrically For any complex numbers z and w you can find the sum z+w by attaching the vector w to the tip of the vector z and completing a paralellogram. The diagonal of that paralellogram is the vector z+w.
To find the difference z-w, reverse the vector w and find z+(-w).
Adding vectors z and w
Subtracting vectors z and w
Multiplication is more difficult to figure out, but feel free to come up with a conjecture before reading on.
Example 1.2.8: Conjecture for multiplying geometrically Let z=1+i, w=1-i, and v=i. Compute and visualize z*w, v*z, v*w, z2, w2, and v2. Try to come up with a conjecture how to visualize multiplication of two general complex numbers.
To see if your conjecture about multiplication is correct, we first need to introduce a more suitable coordinate system. Similar to vectors in the 2-D plane, we define polar coordinates as follows:
Definition 1.2.9: Polar coordinates and cis A complex, non-zero number z=x+iy with rectangular coordinates (x,y) can also be represented in polar coordinates (r,t). Letr = |z| =and t be an angle such thatx = r cos(t) und y = r sin(t)Thenz = x+iy = r cos(t) + i r sin(t) = r (cos(t)+i sin(t)) = r cis(t)where we define cis(t) = cos(t) + i sin(t). r is called the radius or length, t the angle or argument of z.
(x,y) rectangular (r,t) polar (1,0)=1 (1,0)=1 (0,1)=i (1,/2)=cis(/2) (-1,0)=-1 (1,)=cis() (0,-1)=-i (1,3/2)=cis(3/2) (1,1)=1+i (,/4)=cis(/4) (1,-1)=1-i (,-/4)=cis(-/4) Polar coordinates of a complex number Sample points
It is easy to convert a number z from polar coordinates (r,t) to rectangular coordinates (x,y): simply use the fact that z = r cis(t) = r(cos(t) + i sin(t)), i.e. compute
You can verify, for example, that the complex number with polar coordinates r=2 and angle t=/4 can be written in rectangular coordinates as
z = 2 cis(/4) = 2 (cos(/4) + i sin(/4)) = 2 (1/ + i 1/) = + i
Going the other way, i.e. converting from rectangular to polar coordinates, is a little more confusing since the angle in the above definition is not unique. For example, the complex number 1+i clearly has a radius of but any angle /4 + 2k will work. In other words, z=1+i=(1,1) in rectangular coordinates is equal to
(,/4) = (,9/4) = (,17/4) = ...
in polar coordinates. To avoid confusion, we will often (somewhat arbitrarily)
restrict the angle of a complex number in polar coordinates to be between
Definition 1.2.10: The Principle Argument Arg For any complex number z 0 we define the principle argument or Arg(z) as the angle which the vector z makes with the positive (real) x-axis and for which- < Arg(z)
In other words, a non-zero complex number has many arguments, but only one principle argument. To find a principle argument we use the fact that
tan(t) = sin(t)/cos(t) = r sin(t)/r cos(t) = y/x = Im(z)/Re(z)
so that we can use the tan-1 = arctan adjusted in such a way as to give us the principle angle: take any non-zero complex number z:
If Re(z) = 0 then
If Re(z) 0 compute
and adjust the angle as follows:
Example 1.2.11: Polar coordinates examples Convert the following numbers into the indicated coordinates and draw them in the complex plane:
Using polar coordinates will finally allow us to understand how multiplication of complex numbers works geometrically:
Theorem 1.2.12: Multiplying complex numbers geometrically If z = r1 cis(s) and w = r2 cis(t) in polar coordinates thenz*w = r1 r2 cis(s + t)andzn = rn cis(ns)
You might wonder why this theorem is called "multiply complex numbers geometrically" when in fact it shows only formulas. But what the formulas mean is:
Multiplying vectors z and w
A similar trick works for division of two complex numbers:
Corollary 1.2.13: Dividing complex numbers geometrically If z = r1 cis(s) and w = r2 cis(t) in polar coordinates thenz/w = r1 / r2 cis(s - t)
Here are some examples that should convince you that in cases where the angle of a complex number is easy to determine, multiplication and division in polar coordinates can be quite simple.
Example 1.2.14: Multiplying geometrically
- Explain geometrically why (1+i)(1-i), i2, and (1+i)4 are all purely real numbers, i.e. the imaginary part of each answer will be zero.
- Draw the vectors zn, n=1, 2, ... 8 for z=cis(/4) and z=0.85 cis(/8)
- Find both geometrically and algebraically and confirm that both answers agree.
- Compute in your head, then confirm your answer.
If a complex number z lies on the unit circle it has length one so that z=sin(t)+i cos(t) for somes t (i.e. the radius r is 1. In that case the above theorem has another well-known corollary:
Corollary 1.2.15: DeMoivre's Formula For any integer n and any real number t we have(cos(t) + i sin(t))n = cos(nt) + i sin(nt)
DeMoivre's Formula is quite something. It says that if you take a number on the unit circle (i.e. with lenght 1) with initial argument (angle) t and multiply it by itself, it simply rotates around the unit circle by that angle t. Each time you multiply the number by itself, the vector rotates another t degrees. In other words, in this case the power operator results in a simple rotation.
Powers of a vector z with |z|=1
Two interesting questions related to this rotation, taken from the field of Complex Dynamics, are: suppose z is a complex number with |z|=1. Then:
Proposition 1.2.16: Finding Roots For any positive integer n and any non-zero complex number a = r cis(t) the equation zn = a has exactly n distinct roots given by:z =where k = 0, 1, 2, ... n-1.
In a previous example we found the two square roots of i, which turned out to be a fair amount of work. The above proposition allows us to dispense with such a question quickly. For example, the two solutions for
z2 = i = cis(/2)are:
z1 = cis(/2/2) = cis(/4)
z2 = cis((/2 + 2)/2) = cis(5/4)
which you can quickly check using DeMoivre's Formula. Here is a geometric interpretation of this proposition: let's find, for example, the three third-roots of i, i.e. we want to find all solutions to z3 = i.
1: Draw the vector i 2: Divide angle by 3 for first root 3: Draw 3 equally spaced segments,
starting at the first root
Note, in particular, that the third root of i turned out to be -i, which indeed checks out:
(-i)(-i)(-i) = i*i*(-i) = (-1)*(-i) = i
This proposition is very satisfying: it says that at least every simple polynomial equation of degree n has n solutions. Later we will see that this is true in general: every n-th degree polynomial has n roots, no "if's" and "but's". This, in fact, is a sign of things to come: many theorems in complex analysis will turn out to be very "satisfying" and nicely structured, which is one reason that the study of complex analysis is a lot of fun (I think -:). But first a few more 'profane' examples.
Example 1.2.17: Finding roots geometrically
- Find the cube roots of 8 and i and draw them.
- Find all 4 fourth-roots of -1 and draw them geometrically
- Find all 5 fifth-roots of 1 and draw them geometrically
- Find both square-roots of 3i-2 by (a) using polar coordinates and (b) using rectangular coordinates and a formula from the previous section. Confirm that both methods result in the same answers.
Let's conclude this chapter with a result that illustrates what a 'nicely structured' theorem in complex analysis can look like.
Proposition 1.2.18: Roots of Unity The n n-th roots of unity are given by wnk, where k = 0, 1, 2, ... n-1 andwn = cis(2/n)They form the vertices of a regular polygon and add up to zero, i.e. they satisfy the equation:1 + wn + wn2 + ... + wnn-1 = 0
This is neat: not only does the equation zn = 1 have exactly n solutions (one of which is, of course, z=1), but the solutions have this really pretty geometric structure of forming a regular polygon, which implies algebraically that they add up to zero as vectors. Here are, for example, the eight roots of z8=1:
You can see their regular structure. When you add them all as vectors, you indeed get the zero vector.
We will explore some more geometric structures in the next section.