Introduction to Complex Numbers (Edexcel IAL Further Maths)

Cartesian Form of Complex Numbers

Spec point — 1.1

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

  • $x=\pm3$ are real numbers, but neither work

    • They give +9

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

  • $i^{2}=-1$

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

• Rules for surds and indices can be used

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

    • The imaginary solutions are $x=\pm3i$

What is a complex number?

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

  • For example, $3+4i$

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

• This is called Cartesian form

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

  • $z=x+yi$
  • $\text{Re} ( z ) = x$
  • $\text{Im} ( z ) = y$

How are real numbers and complex numbers related?

• The letter $ℂ$ stands for all complex numbers

  • $z\inℂ$

• Complex numbers with no imaginary parts are real numbers

  • The real numbers, $ℝ$, are a subset of the complex numbers, $ℂ$

    • $ℝ\subsetℂ$

• In Cartesian form $z=x+yi\inℂ$

  • $x\inℝ$

  • $y\inℝ$ as $y$ itself takes real values

    • Multiplying it by $i$ makes it imaginary: $yi$

• 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

  • $\begin{gathered} ( 3 + 4 \mathrm{i} ) + ( 2 + 8 \mathrm{i} ) & = ( 3 + 2 ) + ( 4 + 8 ) \mathrm{i} = 5 + 12 \mathrm{i} \end{gathered}$
  • $\begin{gathered} ( 3 + 4 \mathrm{i} ) - ( 2 + 8 \mathrm{i} ) & = ( 3 - 2 ) + ( 4 - 8 ) \mathrm{i} = 1 - 4 \mathrm{i} \end{gathered}$

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

• Multiply or divide their real parts and imaginary parts separately

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

  • $\begin{gathered} ( 3 + 4 \mathrm{i} ) \div 10 & = 0.3 + 0.4 \mathrm{i} \end{gathered}$

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

Exam Hint

• Avoid these handwriting misinterpretations when writing $i$ in the exam:

  • $2\sqrt{5}i$ can look like $2\sqrt{5i}$

    • Alternatives are $( 2 \sqrt{5} ) \mathrm{i}$ or $2i\sqrt{5}$

  • $\frac{3}{2}i$ can look like $\frac{3}{2i}$

    • An alternative is $\frac{3i}{2}$

Worked Example Two complex numbers are given by $z_{1}=p+2i$ and $z_{2}=-7+qi$, where $p$ and $q$ are real.

Given that $z_{1}+2z_{2}=4-8i$, find $p$ and $q$.

Substitute the complex numbers into the left-hand side

$( p + 2 \mathrm{i} ) + 2 ( - 7 + q \mathrm{i} )$

Expand the brackets and collect real and imaginary terms

$\begin{gathered} = p + 2 \mathrm{i} - 14 + 2 q \mathrm{i} \\ = ( \mathrm{p} - 14 ) + ( 2 + 2 q ) \mathrm{i} \end{gathered}$

Compare this to the right-hand side Set the real parts equal to each other and solve

$\begin{gathered} p - 14 & = 4 \\ p = 18 \end{gathered}$

Set the imaginary parts equal to each other and solve

$\begin{gathered} 2 + 2 q & = - 8 \\ 2 q & = - 10 \\ q = - 5 \end{gathered}$

Final answer: $p=18$and$q=-5$

Multiplying Complex Numbers

Spec point — 1.2

How do I multiply complex numbers?

• All rules of expanding brackets still work

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

• For example, $( a + b \mathrm{i} ) ( c + d \mathrm{i} ) = a c + a d \mathrm{i} + b c \mathrm{i} + b d \mathrm{i}^{2}$

  • Use $i^{2}=-1$ in the last term

    • $ac+adi+bci-bd$

  • Then group real and imaginary parts

    • Factorise out the $i$
    • $a c - b d + ( a d + b c ) \mathrm{i}$

• Note that the difference between two squares becomes

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

How do I find powers of i?

• Use the fact that $i^{2}=-1$

  • Below are the first few powers of $i$:

    • $i^{0}=1$
    • $i^{1}=i$
    • $i^{2}=-1$
    • $i^{3}=-i$ from $\mathrm{i}^{2} \times \mathrm{i} = ( - 1 ) \times \mathrm{i}$
    • $i^{4}=1$ from $( \mathrm{i}^{2} )^{2} = ( - 1 )^{2} = 1$
    • $i^{5}=i$ from $i^{5}=(i^{2})^{2}\times i=i$

  • The pattern above continues

    • $i^{6}=-1$
    • $i^{7}=-i$

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

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

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

Exam Hint Questions that say "show your working clearly" won't accept answers written down from a calculator.

Worked Example Showing your working clearly, find and simplify:

(a) $( 4 + \mathrm{i} ) ( 2 + 9 \mathrm{i} )$

Expand the brackets

$=4\times2+4\times9i+i\times2+i\times9i\,=8+36i+2i+9i^{2}$

Collect the imaginary parts Use that $i^{2}=-1$ then collect the real parts

$\begin{gathered} = 8 + 38 \mathrm{i} + 9 \times ( - 1 ) \\ = 8 + 38 \mathrm{i} - 9 \end{gathered}$

$-1+38i$

(b) $( 3 - 4 \mathrm{i} )^{2}$

Write using double brackets then expand

$= ( 3 - 4 \mathrm{i} ) ( 3 - 4 \mathrm{i} ) = 9 - 12 \mathrm{i} - 12 \mathrm{i} + 16 \mathrm{i}^{2}$

Collect the imaginary parts Use that $i^{2}=-1$ then collect the real parts

$\begin{gathered} = 9 - 24 \mathrm{i} + 16 \times ( - 1 ) \\ = 9 - 24 \mathrm{i} - 16 \end{gathered}$

$\begin{gathered} - 7 - 24 \mathrm{i} \end{gathered}$

(c) $( - 2 \mathrm{i} )^{11}$

Use index laws to move the power onto the individual terms

$( - 2 )^{11} \times \mathrm{i}^{11}$

Work out $( - 2 )^{11}$

$( - 2 )^{11} = - 2048$

Work out $i^{11}$ It helps to write it in terms of $i^{10}$ then $i^{2}$ Use $i^{2}=-1$

$\begin{gathered} \mathrm{i}^{11} & = \mathrm{i}^{10} \times \mathrm{i} \\ = ( \mathrm{i}^{2} )^{5} \times \mathrm{i} \\ = ( - 1 )^{5} \times \mathrm{i} \\ = - 1 \times \mathrm{i} \\ = - \mathrm{i} \end{gathered}$

Multiply both parts together Two minus signs make a plus

$- 2048 \times ( - \mathrm{i} )$

$2048i$

Complex Conjugates & Division

Spec point — 1.2

What is a complex conjugate?

• If $z=x+yi$ then the complex conjugate of $z$ is $z\ast=x-yi$

  • The sign of the imaginary part changes

• Note that

  • $z+z\ast$ is always real

    • since $x+yi+x-yi=2x$

  • $z-z\ast$ is always imaginary

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

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

    • since $( x + y \mathrm{i} ) ( x - y \mathrm{i} ) = x^{2} - y^{2} \mathrm{i}^{2} = x^{2} + y^{2}$
    • $zz\ast=z\ast z$

How do I divide complex numbers?

• To divide $z_{1}$ by $z_{2}$, multiply top and bottom of $\frac{z_{1}}{z_{2}}$ by $z_{2}\ast$

  • $z_{2}\ast$ 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, $x+yi$

• The process is called realising the denominator

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

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

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

    • It helps to write the brackets in

  • then expand and simplify

Exam Hint To check your answer in an exam, multiply it by the denominator and see if you get the numerator.

Worked Example Let $z_{1}=1+7i$ and $z_{2}=3-i$.

Find and simplify $\frac{z_{1}}{z_{2}}$, giving your answer in the form $x+yi$ where $x$ and $y$ are real numbers.

Find the complex conjugate of the denominator

$z_{2}\ast=3+i$

Multiply the top and bottom of $\frac{z_{1}}{z_{2}}$ by $z_{2}\ast$

$\frac{( 1 + 7 \mathrm{i} )}{( 3 - \mathrm{i} )} \times \frac{( 3 + \mathrm{i} )}{( 3 + \mathrm{i} )}$

Write as one single fraction then expand top and bottom separately

$\frac{( 1 + 7 \mathrm{i} ) ( 3 + \mathrm{i} )}{( 3 - \mathrm{i} ) ( 3 + \mathrm{i} )} = \frac{3 + \mathrm{i} + 21 \mathrm{i} + 7 \mathrm{i}^{2}}{9 - \mathrm{i}^{2}}$

Use that $i^{2}=-1$then collect real and imaginary parts

$\begin{gathered} = \frac{3 + \mathrm{i} + 21 \mathrm{i} - 7}{9 - ( - 1 )} \\ = \frac{- 4 + 22 \mathrm{i} }{10} \end{gathered}$

To give your answer in the form $x+yi$, split the fraction then simplify

$-\frac{4}{10}+\frac{22}{10}i$

$-\frac{2}{5}+\frac{11}{5}i$

$-0.4+2.2i$is also accepted