A-level Mathematics/Edexcel/Further 1/Complex Numbers

Complex numbers were first developed in the mid 16th Century, as a means for solving certain cubic equations. They consist of an imaginary part (in terms of i, or ${\sqrt {-1}}$) and a real part ('traditional' numbers, so to speak - numbers such as 1, -324 or ${\sqrt {67}}$). Since then, they have become a frequently used type of number in solving polynomial equations and in, unusually, calculations by engineers. In FP1, we consider fairly basic uses of Complex Numbers in solving quartic, cubic and quadratic equations. Similarly, we consider basic arithmetic of complex numbers - adding, subtracting, multiplying and dividing complex numbers.

An Introduction to Complex Numbers

Imaginary Numbers

Previously, one has been unable to solve any equation involving the root of a negative number. For instance, the quadratic equation $y=x^{2}+x+2$ could not be solved as:

 It cannot be factorised
 It cannot be solved with the quadratic equation (as the discriminate would be negative, and one could not square root a negative number).

However, using 'imaginary' numbers allows for negative numbers to be square-rooted:

${\sqrt {-1}}=i$

Applying this, and knowledge of surds, we can use this for any negative number. Thus:

${\sqrt {-9}}={\sqrt {9}}\times {\sqrt {-1}}=\pm 3i$
${\sqrt {-10}}={\sqrt {10}}\times {\sqrt {-1}}=\pm {\sqrt {10}}i$
${\sqrt {-8}}={\sqrt {8}}\times {\sqrt {-1}}=\pm 2{\sqrt {2}}i$

Complex Numbers

A Complex Number is a number containing an imaginary part, in other words something containing $i$, and a real part. So, examples of complex numbers include:

${\mathit {z}}=1+3{\mathit {i}}\!$
$z={\frac {\sqrt {3}}{2}}+3i$
$z=-2+{\sqrt {2}}i$
$z={\frac {3+{\sqrt {5}}i}{2}}$

Even in more complicated examples, one should always be able to separate the real part from the imaginary part. Consider the last example above:

$z={\frac {3+{\sqrt {5}}i}{2}}$

In this number, the real part is $3 \over 2$ and the imaginary part is ${\frac {\sqrt {5}}{2}}i$.

Similarly, if we were to consider:

$z={\sqrt {3}}({\sqrt {2}}+i)$

The real part would be ${\sqrt {3}}\times {\sqrt {2}}={\sqrt {6}}$ and the imaginary part would be ${\sqrt {3}}i$.

One should also take care here: in cases such as ${\sqrt {2}}i$, ${\mathit {i}}\!$ is not under the square root. This represents ${\sqrt {2}}\times i$.

Each of the above examples (and all complex numbers) can be represented by the following:

${\mathit {z}}={\mathit {a}}+{\mathit {b}}{\mathit {i}}\!$ where $a\in \mathbb {R} \!$, $b\in \mathbb {R} \!$.

It is worth familiarising oneself with notation such as $a\in \mathbb {R} \!$. It means a 'is an element of' $\mathbb {R} \!$, where $\mathbb {R} \!$ represents all of the real numbers. This notation in particular is frequently used in FP1.

Complex Conjugate

Each Complex Number has a 'Complex Conjugate'. Despite it's rather complicated name, it is a very simple conecpt:

The Complex Number ${\mathit {z}}={\mathit {a}}+{\mathit {b}}{\mathit {i}}\!$ has the Complex Conjugate ${\mathit {z}}*={\mathit {a}}-{\mathit {b}}{\mathit {i}}\!$.

The Complex Conjugate has multiple uses. For instance, if an equation has the complex root (1+3i), the complex conjugate (1-3i) will also be a root. Similarly, in order to divide complex numbers, one must use the complex conjugate, as we will later see, to 'rationalise' the denominator. One does this by multiplying the fraction by ${\frac {z*}{z*}}$ (which has the same effect, value wise, of multiplying by one). This then allows for the quotient to be re-written in the form a+ib.

For any complex number, the complex conjugate is denoted with an *. For instance, the complex number 'a' has the complex conjugate 'a*'. The complex number 'y' has the complex number 'y*'. Mostly, the examination will use the complex numbers z and w.

Arithmetic with Complex Numbers

Addition of Complex Numbers

Say one was to define:

$z=3+{\sqrt {2}}i$
$w=-2+3{\sqrt {2}}i$

Complex numbers are added together in their separate parts - so, the real parts of the number are added together and the imaginary parts are added. Using our earlier defined examples:

${\begin{aligned}z+w&=(3+{\sqrt {2}}i)+(-2+3{\sqrt {2}}i)\\&=(3-2)+({\sqrt {2}}+3{\sqrt {2}})i\\&=1+4{\sqrt {2}}i\\\end{aligned}}$

Addition of Complex Numbers, just as with real numbers, is the same, regardless of the order in which terms are added - in other words, $z+w\equiv w+z$.

Subtraction of Complex Numbers

Again, if one considers the earlier defined pair of complex numbers, they can be subtracted in the same way.

${\begin{aligned}z-w&=(3+{\sqrt {2}}i)-(-2+3{\sqrt {2}}i)\\&=(3-(-2))+({\sqrt {2}}-3{\sqrt {2}})i\\&=5-2{\sqrt {2}}i\\\end{aligned}}$

As with subtraction of real numbers, subtraction of Complex Numbers is not the same regardless of the order in which the terms are added - in other words:

$z-w\neq w-z$ (provided, of course, both numbers are not zero)

Thus

${\begin{aligned}w-z&=(-2+3{\sqrt {2}}i)-(3+{\sqrt {2}}i)\\&=(-2-3)+(3{\sqrt {2}}-{\sqrt {2}})i\\&=-5+2{\sqrt {2}}i\\\end{aligned}}$

Multiplication of Complex Numbers

Multiplying a Complex Number by a constant is fairly straight-forward - for a constant 'k', you simply multiply the real part by k and the imaginary part by k. So, using our z from earlier:

$kz=k(3+{\sqrt {2}}i)=3k+k{\sqrt {2}}i$

If one was to multiply a complex number by another complex number, one follows an process similar to expanding products such as $({\sqrt {2}}x+2)(x+3)$:

${\begin{aligned}z^{2}&=z\times z\\&=(3+{\sqrt {2}}i)(3+{\sqrt {2}}i)\\&=3\times 3+3\times {\sqrt {2}}i+3\times {\sqrt {2}}i+{\sqrt {2}}i\times {\sqrt {2}}i\\&=9+6{\sqrt {2}}i+2i^{2}\\\end{aligned}}$

It is tempting to leave the solution as one would when multiplying unknowns. However, the value of $i^{2}\!$ is known to be -1. So:

${\begin{aligned}z^{2}&=9+6{\sqrt {2}}i+2i^{2}\\&=9+6{\sqrt {2}}+2\times (-1)\\&=9-2+6{\sqrt {2}}=7+6{\sqrt {2}}\\\end{aligned}}$

Similarly, one can multiply different complex numbers:

${\begin{aligned}z\times w&=(3+{\sqrt {2}}i)(-2+3{\sqrt {2}}i)\\&=3\times (-2)+3\times 3{\sqrt {2}}i+(-2)\times {\sqrt {2}}i+{\sqrt {2}}i\times 3{\sqrt {2}}i)\\&=-6+7{\sqrt {2}}i+6\times i^{2}\\&=(-6-6)+7{\sqrt {2}}i=-12+7{\sqrt {2}}i\\\end{aligned}}$

Division of Complex Numbers

This is a slightly more complex process, similar to rationalising the denominator in situations involving surds.

For instance: ${\frac {2+{\sqrt {3}}i}{{\frac {1}{2}}+{\sqrt {5}}i}}$. In order to 'divide' the top complex number (on the numerator) by the bottom complex number (on the denominator), one must first make the denominator into a real number. In order to do this, one must use the 'complex conjugate' of the denominator. This is, as already seen in section 2, similar to the denominator, only the imaginary part of the number has the opposite polarity. So, in our example, the complex conjugate of ${\frac {1}{2}}+5i$ would be ${\frac {1}{2}}-5i$. Since, however, one cannot simply introduce terms to the fraction. If, however, we use the fraction ${\frac {{\frac {1}{2}}-5i}{{\frac {1}{2}}-5i}}$, we know that this has value one. Yet, multiplying our original fraction by this will change it, allowing for a simplification to take place.

This can by mathematically solved:

$z=2+{\sqrt {3}}i$


$w={\frac {1}{2}}+5i$


$w*={\frac {1}{2}}-5i$

${\begin{aligned}{\frac {z}{w}}&={\frac {2+{\sqrt {3}}i}{{\frac {1}{2}}+5i}}\\&={\frac {2+{\sqrt {3}}i}{{\frac {1}{2}}+5i}}\times {\frac {{\frac {1}{2}}-5i}{{\frac {1}{2}}-5i}}\\&={\frac {(2+{\sqrt {3}}i)({\frac {1}{2}}-5i)}{({\frac {1}{2}}+5i)({\frac {1}{2}}-5i)}}\\&={\frac {2\times {\frac {1}{2}}+2\times (-5i)+{\sqrt {3}}i\times {\frac {1}{2}}+{\sqrt {3}}i\times (-5i)}{{\frac {1}{2}}\times {\frac {1}{2}}+{\frac {1}{2}}\times (-5i)+5i\times {\frac {1}{2}}+5i\times (-5i)}}\\&={\frac {1+(-10i)+{\frac {\sqrt {3}}{2}}+(-5{\sqrt {3}}i)}{{\frac {1}{4}}+({\frac {-5}{2}}i)+{\frac {5}{2}}i+(-25i^{2})}}\\&={\frac {(1+{\frac {\sqrt {3}}{2}})+(-10-5{\sqrt {3}})i}{{\frac {1}{4}}+(-25\times (-1))}}\\&={\frac {{\frac {2+{\sqrt {3}}}{2}}+(-10-5{\sqrt {3}})i}{\frac {101}{4}}}\\&={\frac {4}{101}}({\frac {2+{\sqrt {3}}}{2}}+(-10-5{\sqrt {3}})i)\\\end{aligned}}$

Whilst this process was much expanded for clarity, it does perhaps stress the need for care and the need for accuracy when dividing complex numbers. Similarly, whilst the numbers may be a little complex, the process is fairly straightforward. It is also very similar to 'rationalising the denominator' in that one eliminates the 'i' term from the denominator, allowing the denominator to become a factor of the overall expression.

Dealing with ${\mathit {z}}^{2}$

Imagine the complex number:

${\mathit {z}}^{2}=-8+6{\mathit {i}}\!$

How can one then find ${\mathit {z}}\!$ in the form ${\mathit {a}}+{\mathit {b}}{\mathit {i}}\!$?

Let us initially define ${\mathit {z}}\!$ as: ${\mathit {z}}={\mathit {a}}+{\mathit {b}}{\mathit {i}}\!$


${\begin{aligned}{\mathit {z}}^{2}&=({\mathit {a}}+{\mathit {b}}{\mathit {i}})^{2}\\&={\mathit {a}}^{2}+2{\mathit {a}}{\mathit {b}}{\mathit {i}}+{\mathit {i}}^{2}{\mathit {b}}\!\\&=({\mathit {a}}^{2}-{\mathit {b}}^{2})+2{\mathit {a}}{\mathit {b}}{\mathit {i}}\!\\\end{aligned}}$

This form of ${\mathit {z}}^{2}\!$ is particularly useful as it separates the real part $({\mathit {a}}^{2}-{\mathit {b}}^{2}\!)$ of the complex number from the imaginary part $(2{\mathit {a}}{\mathit {b}}{\mathit {i}})\!$.

We can relate the given value of ${\mathit {z}}^{2}\!$ to the expansion and, thus, we can relate co-efficients:
Since $-8+6{\mathit {i}}=({\mathit {a}}^{2}-{\mathit {b}}^{2})+2{\mathit {a}}{\mathit {b}}{\mathit {i}}\!$, we can deduce that:
$({\mathit {a}}^{2}-{\mathit {b}}^{2})=-8\!$
$2{\mathit {a}}{\mathit {b}}=6\!$

Now, it is clear that simultaneous equations, in ${\mathit {a}}\!$ and ${\mathit {b}}\!$ are present and can be solved using techniques learned at GCSE, and in C1.

$2{\mathit {a}}{\mathit {b}}=6\!$


$\therefore {\mathit {a}}{\mathit {b}}=3\!$


$\therefore {\mathit {a}}={\frac {3}{b}}$
Substituting:
$({\frac {3}{b}})^{2}-{b}^{2}=-8\!$


$\therefore {\frac {9}{b^{2}}}-b^{2}=-8$


$\therefore 9-{\mathit {b}}^{4}=-8{\mathit {b}}^{2}\!$


$\therefore {\mathit {b}}^{4}-8{\mathit {b}}^{2}-9=0$

We now have a quadratic in ${\mathit {b}}^{2}\!$ to solve:

${\mathit {b}}^{4}-8{\mathit {b}}^{2}-9=0\!$


$({\mathit {b}}^{2}-9)({\mathit {b}}^{2}+1)=0\!$


$\Rightarrow {\mathit {b}}^{2}=9,-1\!$


However: ${\mathit {b}}\in \Re \!$
$\Rightarrow {\mathit {b}}^{2}=9\!$


$\therefore {\mathit {b}}=\pm 3\!$

We can then substitute these values back into either of our original equations:

$2{\mathit {a}}{\mathit {b}}=6\!$


${\mathit {b}}=3\Rightarrow 3\times 2{\mathit {a}}=6\therefore {\mathit {a}}=1\!$


${\mathit {b}}=-3\Rightarrow -3\times 2{\mathit {b}}=6\therefore {\mathit {a}}=-1\!$

Finally, we have values for ${\mathit {a}}\!$ and ${\mathit {b}}\!$.

When ${\mathit {z}}^{2}=-8+6{\mathit {i}}\!$, $z=\pm (1+3i)$.

Roots of Polynomial Equations Containing i

Quadratic Equations

There are only two scenarios for solutions of Quadratic Equations (that is, those where the highest power of $x\!$ is $x^{2}\!$). If the co-efficients in the quadratic are all real, one can either have two real roots (which also includes a repeated root, as in the case of $(x+2)^{2}=0\!$) or two complex/imaginary roots. Only equations of this type (with real coefficients) are needed in FP1.

Consider now the quadratic equation ${\mathit {z}}^{2}+1=0\!$. Previously, one would have been unable to solve this as it would require one to root $\pm {\mathit {-1}}\!$. Since we now represent this value as ${\mathit {i}}\!$, this quadratic can now be solved with relative ease:

Using the quadratic formula gives:
${\begin{aligned}z&={-b\pm {\sqrt {b^{2}-4ac}} \over 2a}\\&={-0\pm {\sqrt {0^{2}-4\times 1\times 1}} \over 2}\\&={-0\pm {\sqrt {-4}} \over 2}\\&={\pm {\sqrt {-4}} \over 2}\\&={\pm {\sqrt {4}}{\sqrt {-1}} \over 2}\\&={\pm 2i \over 2}\\&={\pm i}\\\end{aligned}}$

We can then use the factor theorem to prove that either $z=i\!$ or $z=-i\!$ are factors of $z^{2}+1=0\!$.

Let $P(z)=z^{2}+1\!$
${\begin{aligned}z&=i\Rightarrow P(z)=i^{2}+1\!\\&=(-1)+1\!\\&=0\!\\\end{aligned}}$


$P(i)=0\Rightarrow i\!$ is a factor of $z^{2}+1\!$

The same result is achieved for $-i\!$, since ${\sqrt {-1}}=\pm i\!$.

$\Rightarrow P(z)=(z+i)(z-i)$

The same can be done with Complex Roots:

Consider the quadratic equation:
$z^{2}-14z+53\!$


Using the quadratic formula, one can see that:
${\begin{aligned}z&={-b\pm {\sqrt {b^{2}-4ac}} \over 2a}\\&={-(-14)\pm {\sqrt {(-14)^{2}-4\times 1\times 53}} \over 2}\\&={14\pm {\sqrt {196-212}} \over 2}\\&={14\pm {\sqrt {-16}} \over 2}\\&={14\pm 4i \over 2}\\&=7\pm 2i\!\\\end{aligned}}$


If one was to now factorise this expression:


$z^{2}-14z+53=(z-(7+2i))(z-(7-2i))\!$

It should be noted, once again, that $7\pm 2i\!$ is a conjugate pair (since $7+2i\!$ is the complex conjugate of $7-2i\!$ and vice versa). This property is extremly useful, and is often examined in FP1.

Graphical Representations of Complex Numbers

This is the general Argand Diagram, displaying the complex number z = a + bi

This is an interesting concept, when one considers that a part of the complex number is totally imaginary. How can one represent something, in a mathematic manner, that is totally imaginary? An 'Argand Diagram' is the tool used to graphically represent a complex number. It does so by treating the real part of the number as an x-co-orderinate and the imaginary part of the number as a y-co-ordinate.

Argand Diagrams are quite helpful when it comes to determining the 'Argument' of the number. It will also allow for a more visual representation of what is meant by the 'Modulus' of the complex number.

The Modulus of a Complex Numbers

This is very similar to finding the length of any given straight-line on a graph. Imagine that you had two points - the origin (0,0) and point A (3,4). The line OA is drawn, connecting the two points. To find the length of this line, one would use an application of Pythagoras, as learned in C1 and C2:

${\begin{aligned}LengthOA&={\sqrt {(3-0)^{2}+(4-0)^{2}}}\\&={\sqrt {9+16}}\\&={\sqrt {25}}\\&=5\end{aligned}}$
Thus, the line OA has length 5 units