### The complex plane defined

We often use a set of Cartesian-like coordinates to graphically represent complex numbers like

z   =   x   +   iy

with a real part and an imaginary part. Its major elements are illustrated here →.

The horizontal and vertical axes of the graph (which are perpendicular, or "orthogonal") are called the real axis and the imaginary axis, respectively. The are usually labeled Re and Im, as shown. The plane defined with these coordinates is called the complex plane.

Notice that the same transformations between the the Cartesian and polar representations of complex numbers apply as for polar coordinates. It's just simple trigonometry. Also notice in the figure above that the two complex numbers graphed are actually complex conjugates, which we'll generally label z and z*, where (*) stands for complex conjugate. On the complex plane, complex conjugates are reflections of one-another across the real axis.

The polar form of a complex number is:

### z   = r·cos(φ) + ir·sin(φ)

In this section, I will refer to the polar angle as φ, as is commonly done when working on the complex plane.

We will use the term complex number to refer to a number like z = x + iy and a complex vector to be its representation on the complex plane. r is commonly called the modulus of a complex number and φ the angle or the argument.

When we work on the complex plane, we do little to no arithmetic with the imaginary number, i. It just labels the imaginary component of the complex vector; it's like a sign saying "this is the y-coordinate".

### Example 1

#### Converting a complex number, z = x + iy, to polar form

Convert z = 2 + 2i to polar form and plot it on a polar graph.

In this function x = 2 and y (the coordinate along the imaginary axis) is also 2. Remember, the pattern is z = x ± iy. From those coordinates, we can calculate r:

and polar angle, θ:

Now x = r·cos(θ) and y = r·sin(θ), so finally,

Here is a graphical depiction of that coordinate, a vector on the complex plane:

### Example 2

#### Converting a complex number, z = x + iy, to polar form

Convert z = -2 + 2i to polar form and plot it on a polar graph.

This location on the complex plane isn't too different from the first example. The polar radius is the same:

The angle is different, and calculating it is a cautionary tale: Recall that the range of the inverse tangent function is (-π/2, π/2), so it is incapable of directly giving us the angle we need in the upper left quadrant. We have to add π/2 or 90˚ to get the correct angle.

Using x = r·cos(θ) and y = r·sin(θ), we have,

#### Why use the complex plane?

The complex plane allows us to visualize complex numbers geometrically. We can treat them as we do vectors in physics, applying all of the rules of trigonometry to use and manipulate them. On the complex plane, addition of two complex numbers is just normal vector addition—see below. Notice that each of the starting vectors and the sum has a real part and an imaginary part.

### Complex vector multiplication

Multiplication of two complex numbers is also easy to visualize on the complex plane, and it's even simpler to perform algebraically using the polar representation.

Consider two complex numbers, z1 = 1 + i and z2 = 2 + i. These are shown on the right. We know that

z1z2 = (1 + i)(2 + i)

= 2 + i + 2i - 1 = 1 + 3i

The graph shows what the multiplication looks like on the complex plane. Multiplication of two complex vectors results in a vector with a different length, and rotated by the sum of the arguments, φ1 and φ2.

In polar coordinates, the multiplation looks like this:

z1z2 = [r1·cos(φ1) + ir1·sin(φ1)][r2·cos(φ2) + ir2·sin(φ2)]

= r1r2·cos(φ1)cos(φ2) + ir1r2·cos(φ1)sin(φ2) + ir1r2·sin(φ1)cos(φ2) - r1r2·sin(φ1)sin(φ2)

= r1r2·cos(φ1)cos(φ2) - r1r2·sin(φ1)sin(φ2) + ir1r2·cos(φ1)sin(φ2) + ir1r2·sin(φ1)cos(φ2)

= r1r2·cos(φ1 + φ2) + ir1r2·sin(φ1 + φ2)

To get to the last expression, we made use of the trigonometric sum and difference formulae:

cos(a ± b) = cos(a) cos(b) ∓ sin(a) sin(b)

sin(a ± b) = cos(a) sin(b) ± sin(a) cos(b)

#### Vector multiplication in the complex plane

To multiply complex vectors in polar coordinates, multiply the moduli and add the angles:

#### z1z2   =   r1r2·cos(φ1 + φ2) + ir1r2·sin(φ1 + φ2)

A similar formula can be derived for complex vector division in the polar representation. You should derive this one yourself as an exercise.

#### Vector division in the complex plane

To divide complex vectors in polar coordinates, divide the moduli in the same order as the division and subtract the angles (also in that order):

### Multiplication of complex conjugates

You will recall that the complex conjugate of a complex number z = x + iy is z* = x - iy. The product z·z* is (x + iy)(x - iy) = x2 + y2, a real number.

#### The product of a complex number and its complex conjugate is a real number: z·z* ∈ ℜ

Because the product of complex conjugates is real, it must lie on the real axis, with no imaginary part. Often, when it is understood (for example in quantum mechanics) that we're dealing with complex numbers, the expression z2 is taken to mean z·z*.

Now let's take a look at a full example of multiplication of two complex numbers using the polar representation:

### Example 3

#### Calculate the product of z1 = 2 + i and z2 = 3 - 5i

The first thing we need to do to multiply these complex vectors is to convert them to their polar forms. The modulus of z1 is

and the rotation angle is

So the polar form of z1 is

Now the modulus of z2 is

and its angle:

So z2 in polar form is:

Now it's just a matter of multiplying the angles and adding the moduli:

#### Isn't this overly complicated?

You can confirm that our answer in the example above, z1z2 = 11 - 7i, is what you'd get by straight multiplication of the complex numbers in a + bi form. So now you're asking, "why convert to polar form? It seems so complicated." But remember, when we multiply many numbers, that's a lot of binomials and a lot of instances of multiplication of the imaginary number i, and that's a lot of opportunities to make errors. Simply multiplying the moduli and adding the angles is much easier.

### Finding powers of complex numbers: De Moivre's formula

To raise a number to a power, especially a large one, it is much more convenient to work in the polar representation and use De Moivre's formula:

### Proof

To prove De Moivre's formula, begin with the multiplication rule, but multiply one complex number, z, by itself:

zz = r·r·cos(φ+φ) + ir·r·sin(φ+φ)

or     z2 = r2·cos(2φ) + ir2·sin(φ+φ)

Then if we multiply z2 by z, we get:

z3 = r3·cos(3φ) + ir3·sin(3φ)

It is easy to see that the pattern will keep repeating ad infinitum (to infinity) to give:

zn = rn·cos(nφ) + irn·sin(nφ)

This kind of reasoning, building a case that must extend in the same direction to some logical conclusion, is called induction or inductive reasoning.

### Example 4

#### Given that z = 2 + 3i, find z7

First (we're getting used to this), we convert to polar coordinates on the complex plane. The modulus is:

The angle is

The complex vector in polar coordinates is then:

Now we just apply De Moivre's formula to calculate the power:

### Roots of complex numbers

We can use De Moivre's formula, which was an extension of our multiplication rule, to derive a formula for finding the nth roots of a complex number, where n is an integer. We simply change n to 1/n:

That seems fine, but it's not quite finished. The fundamental theorem of algebra tells us that every complex number has n complex nth roots, and this formula would just give us one. To find the others, we have to recognize the cyclic nature of the sine and cosine functions . . .

... they repeat themselves every 2π rad. That means if an angle φ satisfies a trigonometric equation, so must φ + 2π, φ + 4π, and so on.

In our power formula this just didn't matter because φ + 2π is the same angle as φ (we would call these degenerate solutions). But in the root equation above, division by n makes φ + 0π, φ + 2π, φ + 4π, and so on yield different solutions. It looks like this:

Here, k is an integer that starts at zero and counts up. When k is zero, we just have our initial formula for a single root at the beginning of this section.

When k = n, we have this situation:

So our index variable k is only useful up to n-1: k = 0, 1, 2 ... n-1. A couple of examples will make this all a little more clear.

#### nth roots of complex numbers

There are n nth roots of a complex number, which we label ak. They are given by:

where k = 0, 1, 2, ..., n-1

### Practice problems

1. Sketch the complex vector on the complex plane and find its polar representation, z = r·cos(φ) + ir·sin(φ):

a.   4i

b.   -2

c.   √3 + i

d.   7 - 3i

2. Sketch the complex number, z, and also 2z, -z and z/2 on the same complex plane:

a.   z = 1 + i

b.   z = -1 + i√3

3. Sketch z1, z2, z1 + z2 and z1·z2 on the same complex plane:

a.   z1 = 2 - i,     z2 = 2 + i

b.  z1 = -1 + i,     z2 = 2 - 3i

4. Write the complex number in polar form with the angle φ between 0 and 2π

a.   1 + i

b.   1 + i√3

c.   8i

d.   2(i - 1)

5. Find the product, z1·z2 and the quotient, z1/z2:

a.   z1 = cos(π) + i sin(π),
z2 = cos(π/3) + i sin(π/3)

b.  z1 = cos(π/4) + i sin(π/4),
z2 = cos(3π/4) + i sin(3π/4)

6. Find the indicated power using De Moivre's formula:

a.   (1 + i)20

b.   (1 - i√3)5

c.   (2√3 + 2i)5

7. Find the indicated roots and graph them in the complex plane:

a.   the square roots of 4√3 + 4i

b.   the cube roots of 4√3 + 4i

c.   the eighth roots of 1

d.   the eigth roots of -1

8. Convert these complex numbers to polar form: x + iy → r(cosφ + i·sinφ):

a.   -2 + 2i√3

b.   √3 + i

c.   4 - 7i

d.   3 - 3i

9. Convert these complex numbers in polar form to standard form: r(cosφ + i·sinφ)x + iy:

a.   3·[cos(30˚) - i sin(30˚)]

b.   √2·[cos(7π/6) + i sin(7π/6)]

c.   8·[cos(210˚) - i sin(210˚)]

d.   √7·[cos(π/12) + i sin(π/12)]

10. Find the product of z1 and z2, and convert the answer to standard form: x + iy:

a.   z1 = 7[cos(25˚) + i sin(25˚)],
z2 = 2[cos(130˚) + i sin(130˚)

b.   z1 = √3·[cos(3π/4) + i sin(3π/4)],
z2 = (1/3)·[cos(π/6) + i sin(π/6)]

11. Find the quotient and leave it in polar form:

12. Find the indicated power of the complex number and express your answer in z = x + iy form:

a.   (3 + 4i)18

b.   (1 - i√3)3

c.   [1/2 + i(√3/2)]3

13. Find the indicated roots of the complex number:

a.   (-2 + 2i)1/4

b.   (2i)1/5

14. Solve: x5 = 32;   find all roots.

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