Imaginary numbers are essential for practical applications in Physics, engineering, and stats. In this article, we’re going to provide you with an introduction to complex numbers, explain what they are and how to work with them. You can then test your knowledge on our sample questions.
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The topic Complex Numbers builds upon the existing knowledge of the real number system and involves the investigation and understanding of the imaginary numbers.
This builds on prior knowledge and applications of algebra and geometry to the complex number system.
The study of Complex Numbers can be used to appreciate mathematics in the real world and is used in physics, engineering and statistics.
NESA requires students to be proficient in the following outcomes:
Students should feel confident utilising skills like algebra, trigonometry and geometry using the real number system.
Consider \(x^2-9=0\). Solve for \(x\).
\begin{align*}
x^2&=9\\
x&=±3\\
\end{align*}
Now consider \(x^2+9=0\). Solve for \(x\).
\(x^2=-9\)
Notice that there is no ‘real’ solution in this case. There is now a need to introduce an imaginary number \(i\), where \(i=\sqrt{-1}\).
The solutions of \(x^2+9=0\) will be \(x=±3i\).
An imaginary number is defined to be \(ki\) where \(k\) is a real number.
Solve \(x^2 + 40 =0\)
\begin{align*}
x^2 &=-40\\
x&=± \sqrt{-40}\\
x&=±\sqrt{40}i\\
x&=±2\sqrt{10}i\\
\end{align*}
Addition and Subtraction can be performed in a similar way as real numbers, where \(i\) is treated like a pronumeral.
i) \(5i + 10i = 15i\)
ii) \(5i-10i =-5i\)
Recall \(i=\sqrt{-1}\). Therefore
\begin{align*}
i^2&=-1\\
i^3&=-1 \times i = -i\\
i^4 &= -i \times i = 1\\
\end{align*}
See a pattern?
When two imaginary numbers are multiplied or divided together, it will result in a real number.
\(ai\times bi =abi^2 = -ab\)
\(\frac{ai}{bi}=\frac{a}{b}\)
Where \(a\) and \(b\) are real numbers.
i) \(4i\times 8i = 32i^2=-32\)
ii) \( \frac {16i} {20i}= \frac {4}{5}\)
However, when a real number is divided by an imaginary denominator, the numerator and denominator are both multiplied by \(i\).
This is called realising the denominator. It follows a similar process to rationalising the denominator when there is a surd.
Order of operations still remain the same with imaginary numbers, i.e. multiplication and division before addition and subtraction.
Simplify the following.
i) \(-6i+5i \times 8\)
ii) \(\frac {-2 \ \times \ 7 \ + \ 5i \ \times \ 31}{9i}\)
iii) \(i^{2021}\)
Answers:
i) \(-6i+40i=34i\)
ii) \( \frac {-14-15}{9i} \times \frac {i}{i} = \frac {-29i}{-9} = \frac {29}{9} i \)
iii) We know that the powers of \(i\) cycle in multiples of \(4\), i.e. \(i^4=1\).
Therefore \(i^{2020} = 1\) so, \(i^{2021} = i\)
In the previous examples, all questions were asked solely in the complex number system domain. This can be extended to include both real and imaginary numbers in the same space.
Such complex number is given in the form \(a+bi\) where \(a\) and \(b\) are real numbers.
The complex conjugate of \(z=a+bi\) is \(\bar z= a-bi\)
The addition of \(z\) where \(z=x+yi\) and its conjugate \(\bar z\) is \(2Re(z)\)
The subtraction of \(z\) and its conjugate \(\bar z\) is \(2iIm(z)\)
Simplify the following:
i) \(\overline { (2-i)^4 }\)
ii) \(\frac {5}{3 \ + \ i} (1+i)^2 – \overline {(3+ i)^2}\)
Answers:
i) \(\overline {16-32i+24-8i+1}\\
=\overline{16 -32 i + 24 \ – \ 8i + 1}\\
=\overline{42 \ – \ 40i}\\
= 42 + 40i \)
ii) \( \frac{5}{3 \ + \ i} \times \frac{3 \ – \ i}{3 \ + \ i} \times (1+2i) \ – \overline {(9+6i-1)}\\
= \frac{5(3 \ + \ 6i \ – \ i \ + \ 2)}{10} \ – (8-6i)\\
= \frac {5 \ + \ 5i}{2} – 8 + 6i\\
= \frac {-11 \ + \ 17i}{2} \)
Complex numbers can be expressed as equal to each other. For example, \(a+bi=c+di\) where \(a=c\) and \(b=d\), and \(a, b, c, d\) are real.
To solve these questions, you will need to equate the real and imaginary parts.
Solve for \(x\) and \(y\), where \(x\) and \(y\) are real numbers
i) \(2x+3yi=5-3i\)
ii) \((x+5yi)(3-2i)=6-2i\)
iii) \( \frac {4+3i}{8-i} =5x+yi \)
Answers:
i) \begin{align*}
2x=5 \ \text{and} \ 3yi&=-3i\\
x=\frac{5}{2} \ \text{and} \ y&=-1\\
\end{align*}
ii) \begin{align*}
3x-2xi+15yi+10y &= 6-2i\\
(3x+10y)+i(15y-2x) &= 6-2i\\
\end{align*}
\(\text{Solve simultaneously}\\
x=\frac{22}{13} \ \text{and} \ y=\frac{6}{65}\)
iii) \begin{align*}
\frac{4 \ + \ 3i}{8 \ – \ i} \times \frac{8 \ + \ i}{8 \ + \ i} &= 5x +yi\\
\frac{32 \ + \ 4i \ + \ 24i \ – \ 3}{65} &= 5x + yi\\
\frac{29 \ + \ 28i}{65} &= 5x + yi\\
\frac{29}{65} = 5x \ \text{and} \ \frac{28}{65} &= y\\
x= \frac{29}{325} \ \text{and} \ y&=\frac{28}{65}\\
\end{align*}
To find the square roots of a complex number \(a+ib\)
1. Let \(x+yi= \sqrt{a+bi}\)
2. Square both sides: \((x+yi)^2=a+bi\)
3. Expand both sides: \(x^2+2xyi-y^2=a+bi\)
4. Equate real and imaginary parts: \(x^2-y^2=a, 2xy=b\)
5. Solve these equations simultaneously for \(x\) and \(y\)
Find the square roots of \(3+4i\)
| \begin{align*} \text{Let} \ x+yi &= \sqrt{3+4i}\\ (x+yi)^2&=3+4i\\ x^2+2xyi-y^2&=3+4i\\ x^2-y^2=3 , \ xy&=2\\ \text{Substitute} \ y&= \frac{2}{x} \ \text{into} \ \text{the first equation:}\\ x^2- \frac{4}{x^2} &= 3\\ x^4-3x^2-4&=0\\ (x^2-4)(x^2+1)&=0\\ \text{As} \ x \ \text{cannot be imaginary,} \ x&=±2\\ \text{Therefore} \ \sqrt{3+4i}&=±(2+i) \end{align*} |
Find the square roots of \(8-6i\)
| \begin{align*} \text{Let} \ x+yi &= \sqrt{8-6i}\\ (x+yi)^2 &= 8 -6i\\ x^2+2xyi – y^2&=8-6i\\ x^2-y^2=8, \ xy&=-3\\ \text{As} \ x \ \text{cannot be an imaginary number,} \ x&=±3\\ \text{Therefore} \sqrt{8-6i} &= ±(3-i)\\ \end{align*} |
A quadratic equation is in the general form \(ax^2+bx+c\). As such, there are two ways of factorising a quadratic equation.
You may choose whichever method you are more comfortable with in an exam.
i) Fully factorise \(x^2-6x+10\) using the Quadratic Formula
| \begin{align*} \text{Consider} \ x^2-6x+10&=0\\ x&= \frac {6 \ ± \ \sqrt{36-40}}{2}\\ &= \frac{6 \ ± \ \sqrt{-4}}{2}\\ &= \frac{6 \ ± \ 2i}{2}\\ (x-\frac{6 \ + \ 2i}{2})(x – \frac{6 \ – \ 2i}{2})&=0\\ (x-(3+i))(x-(3-i))&=0\\ \text{Therefore} \ x^2-6x+10 &= (x-(3+i))(x-(3-i))\\ \end{align*} |
ii) Fully factorise \(x^2-6x+10\) using the Completing the Square method
| \begin{align*} \text{Consider} \ x^2-6x+10&=(x-α)^2+k\\ &= (x-3)^2+1\\ &= (x-3)^2-i^2 \text{(using} \ i^2=-1 \text{)}\\ &= (x-3-i)(x-3+i) \ \text{(taking the difference of two squares)}\\ \end{align*} |
Sometimes, the equation may contain non-real coefficients.
Solve \(z^2+iz-1-i=0\)
| \begin{align*} \text{To solve this equation, first find the discriminant.}\\ Δ=-1-4(-1-i)&=-1+4+4i=3+4i\\ \text{Let} \ x+yi&= \sqrt{3+4i}\\ \text{From previous example:} \ x+yi&=±(2+i)\\ \text{Now, applying the quadratic formula to the equation:}\\ \text{Therefore} \ z&= \frac{-i \ ± \ (2+i)}{2}\\ z &= \frac{-i \ + \ 2 \ + \ i}{2}, \frac{-i \ – \ 2 \ – \ i}{2}\\ z &=1 \ \text{or} \ -1-i \\ \end{align*} |
1. Simplify: \(i^{1-4n}\) where \(n\) is an integer
2. Simplify: \(Re(\frac{5 \ + \ i}{3 \ + \ i} \ +(3-2i)^3)\)
3. Solve: \(z^2-3z+1-3i=0\)
| 1. \(\frac{i^1}{i^{4n}} = \frac{i}{1} = i\)
2. \(Re(\frac{16-2i}{10} + 27-54i-36+8i) = \frac{16}{10}+27-36=-\frac{37}{5}\) 3. \begin{align*} Δ=(-3)^2 -4(1-3i) &= 5+12i\\ |
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