Complex Numbers 2

Introduction

The Second place complex numbers appear in the average A level syllabus is in the FP2 module. This is more challenging than the first chapter, but from personal experience the ideas aren't too hard to grasp, there is just a truck load of things to remember. Some questions can be super simple, but if you set off on the wrong path, can take an age.

Solution

Solution

Polar Form

Any point in a two dimensional space can be represented in polar form and not just it’s (x,y) coordinate!

Since complex numbers are just a point in an Argand space (a two dimensional space) we can represent them in polar form. This is also called “Modulus-argument form” as those are the two peices of information we take to represent a point in polar form.


Figure 1- Diagram showing arument (θ) and modulus (r)

How to express a point in Polar form

Our Complex number (shown above) is given by the co-ordinates (x,y) so is equal to x+yj (x is the real part, y is the imaginary part). We can also represent this point as it’s distance from the origin (modulus) and the angle this modulus makes with the positive real axis (argument) .

It is clear from figure 1 that the following will always be true:

As a complex number is given in the form x+yj, replace x and y with the expressions above, which the diagrams shows to be clearly ture:


Skills Check

What is the following complex number in modulus-argument form?


Check Your Answer


Operations in Polar form

Polar form really simplifies operations between two complex number!!

  • For multiplications simply multiply the two modulii together, and add the arguments!
  • For division simply divide the two modulii, and take away the arguments!


Skills Check

Convert these two complex numbers into modulus-argument form and multiply the two together.


Check Your Answer


De Moivre’s theorem

Using the information above about multiplying complex numbers in polar form, and applying it to the square and the cube of a given complex number, a cool pattern arises:


This ends up not just being a peculiar pattern, but a very useful general rule in mathematics. This is called the "De Moivre Theorem" and is expressed below:


Using De Moivre's to simplify trigonometric functions

One of the reasons people began to accept complex numbers in the eighteenth century was because they could be used to obtain solutions involving only real numbers much easier than using conventional methods with only reals.

An example of this is expressing multiple angles of a cosine for example in terms of powers of cosθ and vice versa.


To do the opposite (write powers of trig functions in terms of multiple angles) we will need the following rules:

How are these derived?


De Moivre Problem Sheet

De Moivre Problem Sheet: Solutions

Solutions

Complex Exponents

Complex numbers can be represented in the form x+yj and the polar form discussed above. But one of the most elegant ways to represent complex numbers is in exponential form, but how?


This means that any complex number can be represented in the following way:

Where R is the modulus of the complex number, and θ is it's argument.


Skills check

Check Your Answers

Summation using complex numbers

Some series do not looked like they can be summed using methods learnt in A level maths. They aren't geometric or arithmetic. However using techniques involving complex numbers these series can be summed to infinity, or to an "nth term". Take for example the following series, they aren't arithmetic or geometric:

However if C+jS is summed then a pattern we recognize emerges:

The sum of the two sequences is a series of complex numbers in polar form, re-writing these in exponential form (see previous section) gives us a geometric sequence, which is easy to sum to infinity:

In exponential form it is clear that there is a first term, and a common ratio, therefore we can sum this series to infinity:


The whole objective of this venture was to sum the separate series by first combining them to get something which we can sum. To get back to separate sums for each of the series (C and S) we can manipulate the expression we have to make it clear which parts are real and imaginary. The real parts will be the sum of the C series, and the imaginary parts will be the sum of the S series.

To finally get a sum for each separate series (C and S) we separate the real and imaginary parts of what we have simplified the sum to infinity. For the C sequence we take the real parts, and for the S sequence we take the imaginary parts. We have therefore summed 2 sequences we previously couldn't sum, by using complex numbers, and getting real results.

The same logic (using C+jS) can be used to sum series which aren't infinite (just use the general case for the sum of a geometric series).

In the A level exam you'll always be given both the C and the S sequences, no need to invent anything!

Series Problem Sheet

Series Problem Sheet: Solutions

Complex roots of unity

Every Polynomial of degree "n" has "n" roots. This was known as early as 1629, however was proved by Gauss in 1799. This means even the following simple equation has "n" roots:

One of the roots to an equation in this form is clearly always going to be 1, and if "n" is even -1 would also be a root, but what about the others?

First, we can express the numbers we are looking for (z) in modulus-argument form, taking advantage of De Moivre's theorem:


Expressing 1 in polar form, will help us equate the two sides of our equation:


For two complex numbers to be equal, both their moduli and their arguments must be the same. We know that the modulus of 1 is 1, therefore the modulus of all the nth roots must be 1 also. As for the arguments, equating the argument nθ with the arguments of one gives us the following:

Therefore nθ can take any integer multiple of 2π:

Dividing through by n gives us the argument of our nth roots:

Since there are n roots to the equation, substituting integer values for k from 0 to (n-1) will give us all the nth roots "of unity":

*Notice that 1 will always be a root, no matter which "n value" the equation has.

Illustrating the nth roots of 1 on an Argand diagram gives the following:

Complex roots: general case

Discuss on Forum!
Topic Checklist
More Maths