A-level Mathematics/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 do 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 $i^3=i^2*i^1$

$i^1=i.$

$i^2=-1.$

$Thus,-1*i=-i$

Finally: Template:Helpful hint

$i^4=1$

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 + y i. (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:

$(2+3i)(3+4i)=(2*3)+(2*4i)+(3i*3)+(3i*4i)=6+8i+9i+12i^2=12i^2+17i+6$

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

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When representing a complex number on an Argand diagram, we can see that

$sin\theta =\frac{y}{r}$

$y=rsin\theta$

Likewise:

$x=rcos\theta$

Since a complex number z can be represented by x+iy,

$z=rcos\theta +irsin\theta$

$z=r(cos\theta +isin\theta )$

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.

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 rationalizing 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:

$x^3+4x^2+9x+10=(x-b)(x-(-1+2i))(x-(-1-2i))$.

$=(x-b)(x^2+x+2xi+x+1+2i-2xi-2i-4i^2)$

$=(x-b)(x^2+2x+5)$

$=x^3+2x^2+5x-b{x}^2-2bx-5b$

$=x^3+(2-b)x^2+(5-2b)x-5b$

$\Rightarrow \{\begin{array}{c}(2-b)=4\\ (5-2b)=9\\ -5b=10\end{array}$

$\Rightarrow b=-2$

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 $x^2-2x+5$.

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+iy, we denote the answer by p+iq and equate:

$(p+iq{)}^2=x+iy$

$p^2+2ipq-q^2=x+iy$

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:

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:

$(\frac{4}{2q}{)}^2-q^2=2$

$\Rightarrow \frac{16}{4q^2}-q^2=2$

Multiplying by $4q^2$:

$16-4q^4=8q^2$

$\Rightarrow 4q^4+8q^2-16=0$

$\Rightarrow q^4+2q^2-4=0$

Substituting q² for u:

$u^2+2u-4=0$

Using the quadratic formula:

$u=\frac{-2\pm \sqrt{4-4\cdot 1\cdot -4}}{2}$

$\Rightarrow u=\frac{-2\pm \sqrt{20}}{2}$

Since q is real, it must be equal to the square root of the positive value of u:

$q=\sqrt{\frac{-2+\sqrt{20}}{2}}=\sqrt{-1+\frac{\sqrt{20}}{2}}=\pm 1.11...$

If q=+1.11 then:

$p=\frac{4}{2*1.11}=1.80$

If q=-1.11 then:

$p=\frac{4}{2*-1.11}=-1.80$

Hence p+qi is:

$1.80+1.11i,-1.80-1.11\Rightarrow p+qi=\pm (1.80+1.11i)$