Background: Radians

**Getting used to complex numbers**

What is a number? Difficult to say, huh?

As you progress from pre-school, your concept of number is extended by thinking about number *systems* rather than just numbers as single objects (there is no room in the Science Museum where you can go and see the number 2).

1. Counting (1, 2, 3, 5, 6… how much higher did you get before pre-school)

2. Zero and negative whole numbers

3. Fractions

4. Surds like √2. (That one was slipped in, but it’s actually a big extension of your number system).

Why not extend further? Add another number which is √−1. Just as you have to add numbers like 3+4√2 when you add √2, so also you must add numbers like 3+4√−1 when you add √−1.

To picture √−1 we have to expand from a number *line* to a number *plane*.

Activity: Ex.1D Q.1. Ex.1A Q.1-6.

You might think that you can extend the number system more and more, so that you can get answers to more and more sums, would go on forever. And you can go from numbers in two dimensions (ℂ) to numbers in three dimensions, then four, then five. *You can’t*.

Complex numbers are the end of it. On the one hand, it turns out that in ℂ you can find answers to many more sums that don’t have answers in ℝ than just x^{2}+1=0. In fact you can find answers to *any algebraic equation* (any equation that is just made up of powers of x with coefficients); this result is so important that it is called the Fundamental Theorem of Algebra.

Click here for the topological proof of the Fundamental Theorem of Algebra.

On the other, it can be *proved* that ℂ is the *only* finite extension of ℝ which keeps arithmetic working as it does in ℝ. There are no three-dimensional numbers, for example. See here for why the complex numbers are the “final” number system.

**Multiplying complex numbers**

You multiply complex numbers just like you multiply surds with √2, except that (√2)2=2 and i2= –1

Do these surd multiplications as practice

1. (5+√2)(3+4√2)

2. (6+3√2)(7+2√2)

3. (5−2√2)(1+5√2)

Then do these − Ex.1B p.6, Q.1-6

Example: (5+i)(3+4i) = 15+20i+3i+4i^{2} = 15 + 23i − 4 = 11+23i

Check your answers against the back of the book.

**Dividing complex numbers**

You divide complex numbers just like you divide surds with √2, except that (√2)^{2}=2 and i^{2}= –1

Instead of rationalising the denominator, you’re making the denominator a real number

Do these surd divisions as practice

1. ^{(11+4√2)}⁄_{(3+√2)}

2. ^{(1+√2)}⁄_{(2+√2)}

3. ^{(3−5√2)}⁄_{(1+3√2)}

Now do these in the same way

Ex.1C p.9

5. ^{(11+4i)}⁄_{(3+i)}

6. ^{(1+i)}⁄_{(2+i)}

7. ^{(3−5i)}⁄_{(1+3i)}

8. ^{(3+5i)}⁄_{(6−8i)}

**Faster with dividing complex numbers**

These calculations with complex numbers are actually easier than with surds:

1. Because you have only one square root, √(−1) also known as i, but with surds you could have √2, √3, √7 or whatever all in the same sum

2. Because there is a short−cut to calculate the denominator of the answer. If the expression you are dividing by is (a+bi), then you multiply top and bottom by (a−bi), and the result is a^{2}+b^{2}.

So, in Q.5 it is 3^{2}+1^{2}=10, in Q.6 it is 2^{2}+1^{2}=5

Do these using the short cut to find the denominator of the answer

9. ^{(28−3i)}⁄_{(1−i)}

10. ^{(2+i)}⁄_{(1+4i)}

(Check answers from back of FP1 textbook)

**Conjugates of complex numbers**

a+b√c and a−b√c are called conjugate surds.

In the same way, the conjugate of a complex number z=a+bi is a−bi, called z*.

z+z*=2a, and zz* = a^{2}+b^{2}

Do these:

Ex.1C p.9

1. Write the complex conjugate z* for

a. z=8+2i

b. z=6−5i

c. z=⅔−½i

d. z=√5+i√10

2. Find z+z* and zz* for

a. z=6−3i

b. z=10+5i

c. z=¾−¼i

d. z=√5−3i√5

Geometrically, z* is the reflection of z in the real axis. Draw an Argand diagram and mark in the complex numbers 8+2i and 6−5i and their conjugates.

**The symmetry of complex numbers**

We decided on i as an extra number so that our number system would include a square root of −1.

But then it follows that −i is also a square root of −1.

Suppose two mathematicians, Carl Friedrich and Jean-Robert, are constructing the complex number system. Carl Friedrich chooses i to represent one square root of −1, and so −i represents the other.

Jean-Robert uses j instead of i to represent his first square root of −1. (As electrical engineers do: they don’t use i, because they already use it to represent “electrical current”). But Jean-Robert makes his j equal to Carl Friedrich’s −i, and his −j equal to Carl Friedrich’s i.

Once they have made their choices, Jean-Robert’s diagrams will look upside−down to Carl Friedrich, and vice versa, but there *won’t be any difference* between Carl Friedrich’s and Jean-Robert’s calculations, so long as you swap i for j or j for i.

**The complex number system is symmetrical with respect to reflection in the x (real) axis**. (The number line is

*not*symmetrical. Why? The complex number system is

*not*symmetrical with respect to reflection in the y (imaginary) axis.)

So (zw)* = z*w* and (z+w)*.

And if you have any quadratic or cubic or quartic (or any sort of polynomial) equation with real coefficients, and a+bi is a root, then (a+bi)*, which is a−bi, is also a root.

(Carl Friedrich Gauss was the first mathematician to develop the theory of complex numbers as a number *system*. Jean-Robert Argand came up with the idea of picturing complex numbers as points in a plane, hence “Argand diagrams”).

Activity: write down another root of each of these equations.

1. x^{2}+2x+2=0; one root is −1+i

2. x^{3}−5x^{2}+9x+13=0; one root is 2−3i

3. x^{4}−6x^{3}+24x^{2}+6x−25=0; one root is 3+4i

**Real part and imaginary part**

Complex numbers can be pictured as points in a number *plane*, just like ordinary (“real”) numbers can be pictured as points on a number *line*.

So each complex number needs two ordinary (“real”) numbers to describe it.

The x−coordinate of the complex number a+bi is a, and is called its “real part”, or Re(a+bi)

The y−coordinate of the complex number a+bi is b, and is called its “imaginary part”, or Im(a+bi)

The “imaginary part” of a complex number is a real number. (Rather like the y−coordinate of a point is just an ordinary number, like 5, and not “5 up”. Or the denominator of a fraction, like ¾, is an ordinary whole number, like 4, not ¼).

If two complex numbers are equal, then their real parts are equal, and their imaginary parts are also equal.

Activity: Ex.1G Q.1−5

**Modulus and argument**

If complex numbers are pictured as points in a number plane, then there is another way of describing them, not by distance across plus distance up (x and y), but by distance from the origin plus direction (measured as an angle anticlockwise from the positive x−axis).

The distance is called the *modulus* of the complex number z, or |z|. The direction is called the *argument* of z, or arg(z).

(Question: what units is the distance measured in? Answer: whatever units make the distance between 0 and 1 equal to 1. In the same way as the absolute value of a positive or negative whole number equals its distance from zero, in whatever units make the distance between 0 and 1 equal to 1).

Activity: Draw these complex numbers on polar−and−cartesian graph paper and find their modulus and argument (in degrees) by using a ruler and a protractor.

1+i 1−i −1+i −1−i 1+√3i √3+i i 2i 3 −3

If a complex number z has modulus r and argument θ, then its real part is r cos θ and its imaginary part is r sin θ.

|z| = r, and arg(z) = θ. Re(z)= r cos θ and Im(z)= r sin θ

Activity: draw these complex numbers on polar−and−cartesian graph paper and find their real and imaginary parts by using a ruler.

a. 3√2 (cos π/4 + i sin π/4)

b. 6 (cos 3π/4 + i sin 3π/4)

c. √3 (cos π/3 + i sin π/3)

(as in Ex.1F Q.3: answers in back of textbook).

Neater way of writing things: you can write cos θ + i sin θ as cis θ for short−cut

z=r cis θ means |z|=r and arg(z)= θ

**Modulus and argument and conjugates**

|z*| = |z|

|z|^{2} = zz*

arg(z*) = −arg(z)

Activity: look back at your answers to the last activity, and write down the modulus and argument of each of these numbers

a. 3 − 3i

b. −3√2 − 3√2i

c. (√3/2) − (3/2)i

**Modulus and argument are useful for calculations where you multiply complex numbers, or find powers, or square or cube roots**

If you think of a complex number z in terms of the *operations* required to get to it starting from 1, then |z| tells you how much you must *enlarge*, and arg(z) tells you how much you must *rotate*.

When you multiply enlargements (do one enlargement after another), you multiply the scale factors.

When you multiply rotations (do one rotation after another), you *add* the angles you rotate by.

So if z and w are complex numbers

|zw| = |z| |w| and |z/w| = |z|/|w|

arg [zw] = arg[z] + arg [w] and arg[z/w] = arg[z] − arg [w]

If z = r cis θ, then z^{2} = r^{2} cis 2θ, z^{3} = r^{3} cis 3θ, and so on

√z = z^{½} = r^{½} cis (θ/2) (and there is another square root, too)

cube root of z = z^{⅓} = r^{⅓} cis (θ/3) (and there are two other cube roots, too, but more of that later)

**How to calculate modulus and argument on your calculator**

If z=x+iy, then |z| = √(x^{2}+y^{2}), and arg(z) is defined as the angle θ which has cos θ = x/|z| and sin θ = y/|z|

Use your calculator to calculate |3+4i| and arg(3+4i) like this:

Set your calculator to radians

Press Shift + to get Pol(

Key in 3

Press Shift ) to get a comma

Key in 4

Press =

You get r= 5, θ=0.927295. That means |z|=5, arg(z)=0.927295

If you think the argument is an exact fraction of π radians, or the modulus is an exact square root, then press the Alpha key. Press X= and you get the modulus. Press Alpha then Y= and you get the argument.

Activity: Do Ex.1F Q.1 using Pol( , )

Use the facts that |z/w| = |z|/|w| and arg[z/w] = arg[z] − arg [w] to do Ex.1F Q.2

Use Rec( , ) on your calculator to convert from |z| and arg(z) [or r,θ] to x,y [or Re(z) and Im(z)]

**Solving quadratics, cubics, and quartics using complex numbers**

**Do now:**

**For each of these pairs of roots, write: quadratic equation with those roots (and coefficient of x^{2} equal to one); sum of the roots; product of the roots**

Example: roots 2 and 5

Equation (x−2)(x−5)=0, or x^{2}−7x+10=0.

Sum of roots 7, product of roots 10

1: roots 3 and 4

2: roots −3 and 4

3: roots 1 and 2

4: roots −1 and 1

What rule do you notice?

Roots and factors come to the same thing. If an expression (x−a) is a factor of a quadratic or cubic or quartic f(x), then a is a root: f(a)=0

If a is a root, so f(a)=0, then (x−a) is a factor.

We can prove these facts: Factor Theorem

If a cubic equation has roots a_{1}, a_{2}, a_{3}, then the equation is:

(x−a_{1})(x−a_{2})(x−a_{3})=0

Or, multiplying it out,

x^{3}−(a_{1}+a_{2}+a_{3})x^{2}+(a_{1}a_{2}+a_{2}a_{3}+a_{3}a_{1})x−a_{1}a_{2}a_{3}=0

So, as long as the equation is standardised to make the coefficient of x^{3} equal to 1,

**Sum of roots = minus the coefficient of x^{2}; product of roots = minus the constant term**

With quartics we have four brackets to multiply together, so four minus signs, and:

**Sum of roots = minus the coefficient of x^{3}; product of roots = plus the constant term**.

We can prove these facts by induction.

If you count double roots as two, then with complex numbers every quadratic has two roots, every cubic has three, and every quartic has four.

We know that if p (or a+bi) is a root of a quadratic, cubic, or quartic, then p* (or a−bi) is too (because complex numbers are symmetrical around the real axis).

**How to get all the roots of a cubic or quartic, given one of them**

Example: Ex.1H p.29 Q.5:

5. Show that x=3 is a root of the equation 2x^{3}−4x^{2}−5x−3=0. Hence solve the equation completely.

2×3^{3}−4×3^{2}−5×3−3 = 54−36−15−3=0, therefore 3 is a root

Standardise the equation by dividing through by 2: x^{3}−2x^{2}−(5/2)x−(3/2) = 0

Call the other two roots a±bi. Then sum of roots = 3+a+bi+a−bi=3+2a=2, so a=−½

Product of roots = 3(a+bi)(a−bi) = 3(a^{2}+b^{2}) = 3(¼+b^{2})=(3/2), so b=±½

The other two roots are −½±½i

Activity: Ex.1H Q.4, 6, 7

Ex.1I Q.1, 5, 8

**Different ways to picture complex numbers**

Revised and simplified presentation here:

Picturing complex numbers (docx)

Picturing complex numbers (pdf)

Older presentation below:

**Starter**

(click to download all as pdf)

**Modelling Complex Numbers: 1. Points In The Number Plane; 2. Displacements In The Number Plane**

i is traditionally termed an imaginary number. But it can be modelled, including by things in the real world, as much as “real” numbers can. For example, i represents a phase shift of an alternating electric current by 90^{o}. Do that phase shift again, and the current has shifted by 180^{o}, and has become the opposite (minus of) what it originally was.

Complex numbers can be modelled as well as real numbers, and they are much used in electrical engineering.

First modelling: the number plane

√2 seems less “imaginary” than √(-1), because you can draw a line of length √2. We can draw i, too, as long as we go from a number line into a number *plane*.

Complex numbers are defined by their *position* in the number plane just as real numbers are defined by their position on the number line. x+iy is modelled by the point x units along (to the right) and y units up.

Points in the number plane can be paired off with *displacements* in the number plane (vectors).So we can also model the complex numbers as *displacements* in the number plans.

Practice: Ex.1D Q.4-8

Question from May 2016 FP1

**Modelling Complex Numbers: 3. Enlargements And Rotations In The Plane. Modulus And Argument. 4. Matrices**

You can also think of displacements in the plane as enlargements and rotations, so enlargements and rotations in the plane also model complex numbers.

r∠θ , or r cis θ, means enlarge by a factor r and rotate by θ radians anticlockwise (around the origin)

The real numbers are the operations which are just enlargements (θ=0).

i is a rotation by π/2

r = |z| (or modulus), and θ = arg(z) (or argument).

**Activity**

Use this way of thinking to find the solutions to z^{3}=64. If z=rcisθ, then z^{3}=64 ⇒ r^{3}cis3θ=64 ⇒ r^{3}=64, so r=4, and 3θ=0 or 2π or 4π.

Also do Ex.1H Q.10

**Modelling by matrices**

Enlargements and rotations are also linear transformations, so complex numbers are also modelled by a copy of the subgroup of 2×2 matrices which includes just the matrices showing enlargements and rotations (no shearing, no reflection, no scaling by different amounts in different directions).

The real numbers correspond to the matrices like

3 0

0 3

and the imaginary numbers to matrices like

0 −2

2 0

If the matrix is

r cos θ −r sin θ

r sin θ r cos θ

then the enlargement is r, and the rotation is θ

In this way of thinking, i corresponds to

0 −1

1 0

You add and multiply as you usually add and multiply matrices

Use this way of thinking in matrix questions when asked what geometrical transformation corresponds to e.g.

√3/2 −1/2

1/2 √3/2

Think of the complex number matching the matrix: its cosθ and sinθ are shown by the first column. Use Pol on that. Pol(,) gives θ=π/6, so the matrix (and the complex number) correspond to an anticlockwise rotation of π/6 round the origin.

**Square roots of complex numbers**

Example: find √(3+4i)

Two methods, both better than the book

**One:** via Pol(,)

Pol(3,4) ⇒ 5 cis 0.92729521800161

Square-root modulus, halve argument

⇒ √5 cis 0.46364760900081

Rec(√5,0.46364760900081) = 2 + i

and other square root must be 2 − i

**Two:** sum and difference of squares

Call the square root a+bi

Then |(a+bi)|=√|(3+4i)|

So, squaring, a^{2}+b^{2}=5

Also (a+bi)(a+bi)=3+4i

So, equating real parts, a^{2}−b^{2}=3

Solving those two equations, a^{2}=4 and b^{2}=1

a=±2 and b=±1

If the number you’re finding the square root of has a positive imaginary part, then the signs of the real and imaginary parts of the square roots have to be the same; if it has a negative imaginary part, then the signs of the real and imaginary parts of the square roots have to be opposite.

Therefore the roots are 2+i and −(2+i).

[If we were looking for the square roots of 3−4i, they would be 2−i and −(2−i). The square roots of the conjugate are the conjugates of the square roots].

Activity: Find a square root of each of these numbers (they are the same as in Ex.1G, Q.9-11, only written in modulus−argument form. To check your answers against the back of the book, convert them back into x+iy form, using Rec( ,). Use the fact that if w is a square root of z, then −w is too, to find the second square root. Round your answers to 1 d.p.

9. 25 cis 1.2870

10. 61 cis 1.3895

11. 13 cis −1.1760

Questions covering all the things you have learned about complex numbers. Answers on the second page.

Further tricky questions:

Ex. 1I page 30 Q.6 and 9

Ex. 1G page 23 Q.1, 9, 10

Find all the complex cube roots of 8

Find all the solution of z^{4}=1