Introduction to Complex Numbers (Edexcel A Level Further Maths: Core Pure): Revision Note

Cartesian form of complex numbers

Complex numbers are a set of numbers which contain both a real part and an imaginary part. The set of complex numbers is denoted as $\mathrm{ℂ}$.

What is an imaginary number?

• Up until now, when we have encountered an equation such as $x^2=-1$ we would have stated that there are “no real solutions” as the solutions are $x=\pm \sqrt{-1}$ which are not real numbers

• To solve this issue, mathematicians have defined one of the square roots of negative one as $\mathrm{i}$; an imaginary number

  • $\sqrt{-1}=\mathrm{i}$

  • ${\mathrm{i}}^2=-1$

• We can use the rules for manipulating surds to manipulate imaginary numbers.

• We can do this by rewriting surds to be a multiple of $\sqrt{-1}$ using the fact that $\sqrt{ab}=\sqrt{a}\times \sqrt{b}$

What is a complex number?

• Complex numbers have both a real part and an imaginary part

  • For example: $3+4\mathrm{i}$

  • The real part is 3 and the imaginary part is 4

    • Note that the imaginary part does not include the '$\mathrm{i}$'

• Complex numbers are often denoted by $z$ and we can refer to the real and imaginary parts respectively using $\mathrm{Re}\left(z\right)$and $\mathrm{Im}\left(z\right)$

•  In general:

  • $z=a+b\mathrm{i}$ This is the Cartesian form of z

  • $\mathrm{Re}\left(z\right)=a$

  • $\mathrm{Im}\left(z\right)=b$

• It is important to note that two complex numbers are equal if, and only if, both the real and imaginary parts are identical.

  • For example, $3+2\mathrm{i}$ and $3+3\mathrm{i}$ are not equal

Operations with complex numbers

How do I add and subtract complex numbers?

• When adding and subtracting complex numbers, simplify the real and imaginary parts separately

  • Just like you would when collecting like terms in algebra and surds, or dealing with different components in vectors

  • $\left(a+b\mathrm{i}\right)+\left(c+d\mathrm{i}\right)=\left(a+c\right)+\left(b+d\right)\mathrm{i}$

• Complex numbers can also be multiplied by a constant in the same way as algebraic expressions:

  • $k\left(a+b\mathrm{i}\right)=ka+kb\mathrm{i}$

How do I multiply complex numbers?

• The most important thing to bear in mind when multiplying complex numbers is that ${\mathrm{i}}^2=-1$

• We can still apply our usual rules for multiplying algebraic terms:

  • $a(b+c)=ab+ac$

  • $\left(a+b\right)\left(c+d\right)=ac+ad+bc+bd$

• Sometimes when a question describes multiple complex numbers, the notation $z_1, z_2, \dots$ is used to represent each complex number

How do I deal with higher powers of i?

• Because ${\mathrm{i}}^2=-1$ this can lead to some interesting results for higher powers of i

  • ${\mathbf{i}}^3={\mathbf{i}}^2\times \mathbf{i}=\mathbf{ }-\mathbf{i}$

  • ${\mathbf{i}}^4={\left({\mathbf{i}}^2\right)}^2={\left(-1\right)}^2=1$

  • ${\mathbf{i}}^5={\left({\mathbf{i}}^2\right)}^2\mathbf{ }\times \mathbf{i}=\mathbf{i}$

  • ${\mathbf{i}}^6={\left({\mathbf{i}}^2\right)}^3={\left(-1\right)}^3=\mathbf{ }-1$

• We can use this same approach of using i² to deal with much higher powers

  • ${\mathbf{i}}^{23}={\left({\mathbf{i}}^2\right)}^{11}\times \mathbf{i}={\left(-1\right)}^{11}\times \mathbf{i}=\mathbf{ }-\mathbf{i}$

  • Just remember that -1 raised to an even power is 1 and raised to an odd power is -1

Complex conjugation & division

When dividing complex numbers, we can use the complex conjugate to make the denominator a real number, which makes carrying out the division much easier.

What is a complex conjugate?

• For a given complex number $z=a+b\mathrm{i}$, the complex conjugate of $z$ is denoted as $z^{*}$, where $z^{*}=a-b\mathrm{i}$

• If $z=a-b\mathrm{i}$ then $z^{*}=a+b\mathrm{i}$

• You will find that:

  • $z+z^{*}$ is always real because $(a+b\mathrm{i})+(a-b\mathrm{i})=2a$

    • For example: $(6+5\mathrm{i})+(6-5\mathrm{i})=6+6+5\mathrm{i}-5\mathrm{i}=12$

  • $z-z^{*}$ is always imaginary because $\left(a+b\mathrm{i}\right)-(a-b\mathrm{i})=2b\mathrm{i}$

    • For example: $(6+5\mathrm{i})-(6-5\mathrm{i})=6-6+5\mathrm{i}-(-5\mathrm{i})=10\mathrm{i}$

  • $z\times z^{*}$ is always real because $\left(a+b\mathrm{i}\right)\left(a-b\mathrm{i}\right)=a^2+ab\mathrm{i}-ab\mathrm{i}-b^2{\mathrm{i}}^2=a^2+b^2$ (as ${\mathrm{i}}^2=-1$)

    • For example: $(6+5\mathrm{i})(6-5\mathrm{i})=36+30\mathrm{i}–30\mathrm{i}-25{\mathrm{i}}^2=36–25(-1)=61$

How do I divide complex numbers?

• When we divide complex numbers, we can express the calculation in the form of a fraction, and then start by multiplying the top and bottom by the conjugate of the denominator:

  • $\frac{a+b\mathrm{i}}{c+d\mathrm{i}}= \frac{a+b\mathrm{i}}{c+d\mathrm{i}} \times \frac{c-d\mathrm{i}}{c-d\mathrm{i}}$

• This ensures we are multiplying by 1; so not affecting the overall value

• This gives us a real number as the denominator because we have a complex number multiplied by its conjugate ($z{z}^{*}$)

• This process is very similar to “rationalising the denominator” with surds which you may have studied at GCSE