# A-level Mathematics/OCR/FP1/Complex Numbers
## Imaginary numbers

Before understanding complex numbers, we need to know about the imaginary unit, i. i arises from the need to be able to calculate the square root of negative numbers.
Think of the equation $x^{2}=-1$ . From this, we can see that
    $x={\sqrt {-1}}$
All logic tells you that you cannot multiply anything by itself and achieve a negative number, and this is true. So we give it a value of i.
    $i={\sqrt {-1}}$
From this, we can see that:
    $i^{2}=-1$
From here we can derive that:
    $i^{3}=-i$
As $i^{3}=i^{2}\cdot i^{1}$
    $i^{1}=i$
    $i^{2}=-1$
    Thus, $-1\cdot i=-i$
Finally:
    $i^{4}=1$
As -1 squared is equal to 1.
With this we have a very valuable tool and are now able to do square roots of negative numbers.

## Simple Operations With Complex Numbers

Complex numbers are numbers that consist of a real number and an imaginary number. They come in the form of $x+yi$ . (With $x$ and $y$ being real numbers and $i$ being the imaginary unit).
They add up very much like you would imagine. Take the two complex numbers (2+3i) and (3+4i). Adding the two together is as simple as adding the real parts (in this case: 2 + 3 = 5) and adding the imaginary parts (3i + 4i = 7i) to give us the final complex number of (5 + 7i).
The same applies for subtraction. Taking our example complex numbers of (2+3i) and (3+4i). We just subtract the real part (2 - 3 = -1) and subtract the imaginary part (3i - 4i = -i) to give us our new complex number (-1-i).
Multiplication is much like expanding quadratic and cubic equations. Simply, you just take the two complex numbers, sit them side by side and multiply out. Taking our example complex numbers we would do the following:
$(2+3i)(3+4i)=(2\cdot 3)+(2\cdot 4i)+(3i\cdot 3)+(3i\cdot 4i)=6+8i+9i+12i^{2}=12i^{2}+17i+6$
It really is as simple as thinking of $i$ like $x$ in our quadratics and treating it in exactly the same way, then replacing $i^{2}$ with $-1$ , $i^{3}$ with $-i$ and $i^{4}$ with $1$ .

## Argand diagrams

The modulus of a complex number is the length of the line that joins the origin of an Argand diagram to the point that represents the complex number, and it is given by:
$|z|={\sqrt {z{\bar {z}}}}={\sqrt {x^{2}+y^{2}}}$
$z=x+yi$
The argument of a complex number is the angle in radians of the complex number point from the real axis (the x-axis), measured counterclockwise. The argument for a complex number z, denoted arg(z), is given by:
$\theta =\arctan \left({\tfrac {y}{x}}\right)$
if it is in the first quadrant.
When representing a complex number on an Argand diagram, we can see that
    $\sin(\theta )={\frac {y}{r}}$
    $y=r\sin(\theta )$
Likewise:
    $x=r\cos(\theta )$
Since a complex number z can be represented by x+iy,
    $z=r\cos(\theta )+ir\sin(\theta )$
    $z=r(\cos(\theta )+i\sin(\theta ))$
A complex number can be represented in either Cartesian or polar form. The polar form is:
$z=r(\cos(\theta )+i\sin(\theta ))$
$r$
$|z|$
$\theta$
The polar form of a complex number is very similar to how we represent vectors. Accordingly, the modulus of a complex number is analogous to the resultant of the x and y components, and the argument is the direction of the resultant vector.
Euler's relationship is:
$e^{\theta i}=\cos(\theta )+i\sin(\theta )$
By substituting this into the polar form for a complex number, we can see that another way of writing the complex number z is:
$|z|e^{i\arg(z)}$

## Complex conjugates and equations

Conjugates are just reflections of co-ordinates on an Argand diagram. They are reflected across the real axis and thus only the imaginary co-ordinate is changed (from, say, 2 + 2i to 2 - 2i). In other words, the sign of the imaginary part is changed (from negative to positive or vice versa).
Complex conjugates are useful when dividing complex numbers as the denominator can be made real by multiplying the top and bottom of the fraction by the complex conjugate of the denominator, for example: ${\frac {2+4i}{1+i}}={\frac {(2+4i)(1-i)}{(1+i)(1-i)}}={\frac {2+2i+4}{1+1}}={\frac {6+2i}{2}}=3+i$
This is called realising (the complex number version of rationalising) the denominator.
Conjugates are also useful when solving equations with real coefficients. If such an equation has an complex root, then the conjugate of the complex number will also be a root of the equation, allowing you to factorise the equation fully in most cases in exams.
Example: Given that $-1+2i$ is a solution, find the other roots of the equation $x^{3}+4x^{2}+9x+10=0$ .
We automatically know one of the other roots is $-1-2i$ (since the coefficients of the polynomial above are all real) so we can start to form a factorization of the equation:
$x^{3}+4x^{2}+9x+10=(x-b)(x-(-1+2i))(x-(-1-2i))$ .
$=(x-b)(x^{2}+x+2xi+x+1+2i-2xi-2i-4i^{2})$
$=(x-b)(x^{2}+2x+5)$
$=x^{3}+2x^{2}+5x-bx^{2}-2bx-5b$
$=x^{3}+(2-b)x^{2}+(5-2b)x-5b$
$\Rightarrow {\begin{cases}(2-b)=4\\(5-2b)=9\\-5b=10\end{cases}}$
$\Rightarrow b=-2$
Now that we know about complex numbers, we can start to solve quadratic equations whose determinant is negative (i.e. quadratics that do not cross or touch the x-axis on a Cartesian graph).
Example: Solve the equation $x^{2}-2x+5=0$ .
Since $b^{2}-4ac=-16$ we know that the equation does not have any real roots. Using the quadratic formula:
    $x={\frac {2\pm {\sqrt {-16}}}{2}}$
    $\Rightarrow x={\frac {2\pm 4i}{2}}$
    $\Rightarrow x=1\pm 2i$

## Square rooting complex numbers

Every complex number has two complex square roots. To find them for the general complex number x+yi, we denote the answer by p+iq and equate:
    $(p+qi)^{2}=x+yi$
    $p^{2}+2ipq-q^{2}=x+yi$
Since x is the only real term on the right-hand side, it must be equal to $p^{2}-q^{2}$ , which are the real terms of the left-hand side. Likewise, y must be equal to 2pq. This is a general rule:
$p=r$
$q=s$
If we were trying to find the square roots of, say, 2+4i, we would have:
    $p^{2}-q^{2}=2$
    $2pq=4$
    $\Rightarrow p={\frac {4}{2q}}$
Substituting:
    $\left({\frac {4}{2q}}\right)^{2}-q^{2}=2$
    $\Rightarrow {\frac {16}{4q^{2}}}-q^{2}=2$
Multiplying by $4q^{2}$ :
    $16-4q^{4}=8q^{2}$
    $\Rightarrow 4q^{4}+8q^{2}-16=0$
    $\Rightarrow q^{4}+2q^{2}-4=0$
Substituting q$^2$ for u:
    $u^{2}+2u-4=0$
Using the quadratic formula:
    $u={\frac {-2\pm {\sqrt {4-4\cdot 1(-4)}}}{2}}$
    $\Rightarrow u={\frac {-2\pm {\sqrt {20}}}{2}}$
Since q is real, it must be equal to the square root of the positive value of u:
    $q={\sqrt {\frac {-2+{\sqrt {20}}}{2}}}={\sqrt {-1+{\frac {2{\sqrt {5}}}{2}}}}=\pm {\sqrt {{\sqrt {5}}-1}}$
If $q={\sqrt {{\sqrt {5}}-1}}$ then:
    $p={\frac {4}{2{\sqrt {{\sqrt {5}}-1}}}}={\sqrt {1+{\sqrt {5}}}}$
If $q=-{\sqrt {{\sqrt {5}}-1}}$ then:
    $p={\frac {4}{2\cdot \left(-{\sqrt {{\sqrt {5}}-1}}\right)}}=-{\sqrt {1+{\sqrt {5}}}}$
Hence p+qi is:
    ${\sqrt {1+{\sqrt {5}}}}+\left({\sqrt {{\sqrt {5}}-1}}\right)i\ ,\ -{\sqrt {1+{\sqrt {5}}}}-\left({\sqrt {{\sqrt {5}}-1}}\right)i\ \Rightarrow \ p+qi=\pm \left({\sqrt {1+{\sqrt {5}}}}+\left({\sqrt {{\sqrt {5}}-1}}\right)i\right)$
This is part of the FP1 (Further Pure Mathematics 1) module of the A-level Mathematics text.