- A-Level Further Maths

# Complex Numbers, Argand Diagrams & Loci

We know from single maths that if the discriminant of a quadratic equation is positive, it has two solutions, if the discriminant = 0, there is one repeated root, and if it is negative there are no roots.

Wrong.

If the discriminant is negative, the quadratic has two **imaginary roots**, given in terms of **i**, the square root of -1:

i = √-1

A **complex number** is a **sum of a real and an imaginary number**, given in the form ** a + b i**, where

*a*is the real number and

*b*the coefficient of i. Often, complex numbers are represented by the letter

*z*.### Adding and Subtracting Complex Numbers

To add or subtract complex numbers, treat the real and imaginary terms separately:

(a+bi) + (c+di) = (a+c) + (b+d)i

A few examples:

**3**+**(2+4i)**= (3+2) + (4i) =**5+4i****(1+2i)**-**4i**= (1) + (2-4)i =**1-2i****(2-2i)**+**(6+i)**= (2+6) + (-2+1)i =**8-i**

### Multiplying Complex Numbers

To multiply an imaginary number by a real constant, just expand the brackets:

k(a+bi) = ka+kbi

For example:

**3(2+4i)**= 3(2) + 3(4i) =**6+12i**

To multiply two imaginary numbers, expand like any quadratic, using the fact that:

i² = -1

For example:

**(2+3i)(3+4i)**= 6 + 8i +9i +12i² = 6 + 17i - 12 =**-6+17i**

### Complex Conjugation

Just like conjugate pairs when rationalising a denominator, the conjugate of a complex number has the opposite sign:

For any complex numberz=a+bi, the conjugate is given asz* =a-bi

The complex conjugate is given an asterisk.

For any complex numberz, the product ofzandz* is a real number

For example, if **z**** = 4+5i:**

**z*******=**4-5i****zz*******=**(4+5i)(4-5i)**= 16 + 20i - 20i -25i² = 16 + 25 =**41**

## Solving Complex Equations

**Complex roots always come in conjugate pairs. **Therefore, if the discriminant of a quadratic is negative, if *z* is one of the roots, *z** will be the other root.

For example, the roots of the equation x² - 6x + 13 = 0 are x = (3+2i) and x = (3-2i)

Often, equations with complex roots will use z instead of x.

The same rule applies to cubic and quartic graphs:

If a cubic graph has just one real root, it will have two complex roots,

*z*and*z**If a quartic graph has two real roots or one real repeated root, then it will also have two complex roots,

*z*and*z**If a quartic graph has no real roots, it will have two complex conjugate pairs as roots:

*z*,*z**,*w*and*w**

All cubic graphs must have a total of three roots, real or complex

All quartic graphs must have a total of four roots, real or complex

## Argand Diagrams

Complex numbers can be represented graphically on an A**rgand diagram**. The x-axis is the real axis, Re, and the y-axis is the imaginary axis, Im. To plot a complex number, treat it as a coordinate:

The complex numberz= (a+bi) has coordinates (a,b) on an Argand diagram

This can be represented as a **position vector**, at an angle, θ, to the positive real axis (generally in radians):

Like any vector, the modulus (the magnitude) can be found using Pythagoras' theorem:

|z| = √(a²+b²)

and tan of the angle is given as:

tan(θ) = b/a

## Modulus-Argument Form

The magnitude of the position vector for a complex number, **r**, is known as the **modulus**, and the angle it makes with the positive real axis, **arg z**, is called the

**argument**. Therefore, complex numbers can also be expressed in

**modulus-argument form:**

z= r(cos(θ) + i sin(θ))

**r**=**|***z*|**arg**=*z***θ**

### Multiplying & Dividing Mod-Arg

## Loci

A locus of points is a collection of points that can be anywhere along a particular line. These can be represented with complex numbers on an Argand diagram.

The distance between two complex numbers,z₁ andz₂, is given as |z₂ -z₁|

If, instead of defining both complex numbers, you just define one and give one general complex number, you get the locus of a **circle**:

|z-z₁| = r is a circle with centrez₁ and radius r

This can also be given as** | z - (a+bi)| = r**, where

**r**is the radius and

**(a, b)**the centre.

You can find the locus of points for a** perpendicular bisector** of a line segment between two points:

The perpendicular bisector of a line segment between two complex numbers,z₁ andz₂, is given as |z-z₁| = |z-z₂|

Using the argument of a complex number, *z*₁, we can express the locus of points for a **half line**. This is a line from, but not including, the fixed point *z*₁ at an angle of arg(*z*-*z*₁):

arg(z-z₁) = θ is a half line from but not including the pointz₁ at an angle of θ to the horizontal.

### Regions

Loci can be used to represent regions on an Argand diagram:

## Exponential Form

Complex numbers can also be expressed in **exponential form**, z = r e^iθ, using **Euler's relation**:

Euler's relation is derived from the Maclaurin series expansions of sin(θ), cos(θ) and eˣ. See Further Maths Notes Sheet on Series.

This leads to the **exponential form for a complex number**,** ***z*:

Where

**r**=**|**and*z*|**arg**=*z***θ**just like in modulus-argument form.

### Multiplying & Dividing

The same rules as for modulus-argument can be applied to multiplying and dividing complex numbers in exponential form:

## De Moivre's Theorem

Euler's relation can be used to find powers of complex numbers that are in the modulus-argument form.

(r (cos(θ) + i sin(θ))ⁿ = rⁿ(cos(nθ) + i sin(nθ))

For any integer

*n*

This can be proven with Euler's relation:

It can also be proven using **proof by induction** (See the Notes Sheet on this)

### Trigonometric Identities

De Moivre's theorem, combined with the binomial expansion, can be used to derive trigonometric identities.

**To drive an equation for sin(nθ) or cos(nθ):**

**To drive an equation for sinⁿ(θ) or cosⁿ(θ):**

There are four standard results you need to know:

An example of how to apply these:

## nth Roots of a Complex Number

Just as a real number, x, has two square roots, ±√x, any complex number has *n* distinct *n*th roots.

Ifzandware non-zero complex numbers andnis a positive integer, the equationzⁿ =whasndistinct roots.

We can use de Moivre's theorem to find the solutions to *z*ⁿ = *w* by taking into account the fact** that its argument is repeated every 2π**.

For any complex numberz= r(cos(θ) + i sin(θ)),z= r(cos(θ + 2kπ) + i sin(θ + 2kπ))

Where k is any integer

### When z**ⁿ** = 1

In general, the solutions tozⁿ = 1 arez= cos(2kπ /n) + i sin(2kπ /n) for k = 1, 2, 3, ...,n

These are known as the *n*th roots of unity.

If *n* is a positive integer, then there is an *n*th root of unity *ω* such that:

the

*n*th roots of unity are 1,*ω*²,*ω*³,*ω*⁴, ...,*ω*ⁿ‾¹1,

*ω, ω*²,*ω*³,*ω*⁴, ...,*ω*ⁿ‾¹ form the vertices of a regular*n*-gon1 +

*ω*² +*ω*³ +*ω*⁴ + ... +*ω*ⁿ‾¹ = 0

### Using *n*th Roots Geometrically

Thenth roots of any complex number,z,form the vertices of a regularn-gon with the origin at its centre

The size and orientation of the polygon depend on the complex number

*z*.

If you know the coordinates of a single vertex, you can find the others by rotating that point about the origin by 2π/*n*. This is the same as multiplying by the *n*th roots of unity.

Ifz₁ is one of the roots of the equationzⁿ =w, and 1,ω, ω²,ω³,ω⁴, ...,ωⁿ‾¹ are the roots of unity, then the roots ofzⁿ =warez₁,z₁ω,z₁ω²,z₁ω³,z₁ω⁴, ...,z₁ωⁿ‾¹