Introduction to Complex Numbers (Edexcel International A Level (IAL) Further Maths: Further Pure 1): Revision Note

Cartesian Form of Complex Numbers

What is an imaginary number?

• Equations like $x^2=-9$ have no real solutions

  • Squaring real numbers always gives a positive value

    • No real number squared could give -9

  • $x=\pm 3$ are real numbers, but neither work

    • They give +9

• To get round this, mathematicians introduce the imaginary number, $\mathrm{i}$, as follows:

  • ${\mathrm{i}}^2=-1$

    • This can be thought of as $\mathrm{i}=\sqrt{-1}$

• Rules for surds and indices can be used

  • $x^2=-9$ means $x^2=9\times \left(-1\right)=9{\mathrm{i}}^2$

    • The imaginary solutions are $x=\pm 3\mathrm{i}$

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

    • The imaginary part is 4

• This is called Cartesian form

• In general, Cartesian form is written using the notation:

  • $z=x+y\mathrm{i}$

  • $\mathrm{Re}\left(z\right)=x$

  • $\mathrm{Im}\left(z\right)=y$

How are real numbers and complex numbers related?

• The letter $\mathrm{ℂ}$ stands for all complex numbers

  • $z\in \mathrm{ℂ}$

• Complex numbers with no imaginary parts are real numbers

  • The real numbers, $\mathrm{ℝ}$, are a subset of the complex numbers, $\mathrm{ℂ}$

    • $\mathrm{ℝ}\subset \mathrm{ℂ}$

• In Cartesian form $z=x+y\mathrm{i}\in \mathrm{ℂ}$

  • $x\in \mathrm{ℝ}$

  • $y\in \mathrm{ℝ}$ as $y$ itself takes real values

    • Multiplying it by $\mathrm{i}$ makes it imaginary: $y\mathrm{i}$

• Complex numbers with no real parts are called imaginary numbers

How do I add or subtract complex numbers?

• Add or subtract their real parts and imaginary parts separately

  • $\begin{array}{rcl}(3+4\mathrm{i})+\left(2+8\mathrm{i}\right)& =& \left(3+2\right)+\left(4+8\right)\mathrm{i}=5+12\mathrm{i}\end{array}$

  • $\begin{array}{rcl}(3+4\mathrm{i})-\left(2+8\mathrm{i}\right)& =& \left(3-2\right)+\left(4-8\right)\mathrm{i}=1-4\mathrm{i}\end{array}$

How do I multiply or divide complex numbers by real numbers?

• Multiply or divide their real parts and imaginary parts separately

  • $\begin{array}{rcl}10(3+4\mathrm{i})& =& 30+40\mathrm{i}\end{array}$

  • $\begin{array}{rcl}(3+4\mathrm{i})÷10& =& 0.3+0.4\mathrm{i}\end{array}$

    • This can also be written $\begin{array}{rcl}\frac{1}{10}\left(3+4\mathrm{i}\right)& =& \frac{3}{10}+\frac{4}{10}\mathrm{i}\end{array}$

Multiplying Complex Numbers

How do I multiply complex numbers?

• All rules of expanding brackets still work

  • You need to remember that ${\mathrm{i}}^2=-1$

• For example, $\left(a+b\mathrm{i}\right)\left(c+d\mathrm{i}\right)=ac+ad\mathrm{i}+bc\mathrm{i}+bd{\mathrm{i}}^2$

  • Use ${\mathrm{i}}^2=-1$ in the last term

    • $ac+ad\mathrm{i}+bc\mathrm{i}-bd$

  • Then group real and imaginary parts

    • Factorise out the $\mathrm{i}$

    • $ac-bd+\left(ad+bc\right)\mathrm{i}$

• Note that the difference between two squares becomes

  • $\left(a+b\mathrm{i}\right)\left(a-b\mathrm{i}\right)=a^2-b^2{\mathrm{i}}^2=a^2+b^2$

How do I find powers of i?

• Use the fact that ${\mathrm{i}}^2=-1$

  • Below are the first few powers of $\mathrm{i}$:

    • ${\mathrm{i}}^0=1$

    • ${\mathrm{i}}^1=\mathrm{i}$

    • ${\mathrm{i}}^2=-1$

    • ${\mathrm{i}}^3=-\mathrm{i}$ from ${\mathrm{i}}^2\times \mathrm{i}=\left(-1\right)\times \mathrm{i}$

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

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

  • The pattern above continues

    • ${\mathrm{i}}^6=-1$

    • ${\mathrm{i}}^7=-\mathrm{i}$

• Find higher powers of $\mathrm{i}$ using a base of ${\mathrm{i}}^2$

  • Remember that -1 to an even power is 1 (or to an odd power is -1)

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

Complex Conjugates & Division

What is a complex conjugate?

• If $z=x+y\mathrm{i}$ then the complex conjugate of $z$ is $z*=x-y\mathrm{i}$

  • The sign of the imaginary part changes

• Note that

  • $z+z*$ is always real

    • since $x+y\mathrm{i}+x-y\mathrm{i}=2x$

  • $z-z*$ is always imaginary

    • since $x+y\mathrm{i}-\left(x-y\mathrm{i}\right)=2y\mathrm{i}$

  • $zz*$ is always real (and non-negative)

    • since $\left(x+y\mathrm{i}\right)\left(x-y\mathrm{i}\right)=x^2-y^2{\mathrm{i}}^2=x^2+y^2$

    • $zz*=z*z$

How do I divide complex numbers?

• To divide $z_1$ by $z_2$, multiply top and bottom of $\frac{z_1}{z_2}$ by $z_2*$

  • $z_2*$ is the complex conjugate of the denominator

• This makes the denominator a real number

  • which allows you to write the final answer in Cartesian form, $x+y\mathrm{i}$

• The process is called realising the denominator

  • It is a very similar to rationalising the denominator with surds

• For example, to work out $\frac{50+75\mathrm{i}}{3+4\mathrm{i}}$

  • calculate $\frac{\left(50+75\mathrm{i}\right)}{\left(3+4\mathrm{i}\right)}\times \frac{\left(3-4\mathrm{i}\right)}{\left(3-4\mathrm{i}\right)}$

    • It helps to write the brackets in

  • then expand and simplify