Complex numbers might seem daunting at first, but they open a whole new dimension in maths. Understanding complex numbers allows you to explore their applications in engineering (like signalling in cellular and wireless technologies) and science (like studying brain waves in biology). Use this resource for an introduction to the basics.
Imaginary numbers
If you have tried to solve algebraic equations, you might notice there are sometimes, there are no "real" solutions. For example, no real number satisfies \(x^{2}+1=0\).
To find solutions to such equations, we use complex numbers. This concept centres around the imaginary number \(i\), where:
\[i=\sqrt{-1}\]
The powers of \(i\) follow a pattern, as shown in the tabke, \(i^{2}=-1\), \(i^{3}=-i\), \(i^{4}=1\), \(i^{5}=i\), and so on.
\(n\)
\(i^{n}\)
\(1\), \(5\), \(9\), \(13\), etc.
\(-1\)
\(2\), \(6\), \(10\), \(14\), etc.
\(-i\)
\(3\), \(7\), \(11\), \(15\), etc.
\(1\)
\(4\), \(8\), \(12\), \(16\), etc.
\(i\)
Complex numbers
A complex number \(z\) is written as:
\(z=x+yi\) where \(x\) and \(y\) are real numbers.
\(x\) is called the real part of \(z\), denoted by \(\Re(z)\) or \(\textrm{Re}(z)\).
\(y\) is called the imaginary part of \(z\), denoted by \(\Im(z)\) or \(\textrm{Im}(z)\).
Two complex numbers are equal only if their real parts are equal and their imaginary parts are equal.
\(a+bi=c+di\) only if \(a=c\) and \(b=d\).
Example 1 – finding solutions for complex numbers
Find the real and imaginary parts of \(z=5-3i\).
Complex numbers are written in the form \(x+yi\). \(x\) is the real part of \(z\), so \(\Re(z)=5\). \(y\) is the imaginary part of \(z\), so \(\Im(z)=-3\).
Find the real and imaginary parts of \(z=\sqrt{3}i\).
\(x\) is the real part of \(z\), so \(\Re{z}=0\). \(y\) is the imaginary part of \(z\), so \(\Im(z)=\sqrt{3}\).
If \(z_{1}=x-\dfrac{i}{3}\), \(z_{2}=\sqrt{2}+yi\) and \(z_{1}=z_{2}\), find the values of \(x\) and \(y\).
\[\begin{align*} \Re(z_{1}) & = \Re(z_{2})\\
x & = \sqrt{2}
\end{align*}\] \[\begin{align*} \Im(z_{1}) & = \Im(z_{2})\\
y & = -\frac{1}{3}
\end{align*}\]
Exercise – finding solutions for complex numbers
Express the following in terms of \(i\) in simplest surd form.
\(\sqrt{-9}\)
\(\sqrt{-2}\)
\(\sqrt{-5}\times\sqrt{3}\)
\(\sqrt{-5}\times\sqrt{10}\)
\(\sqrt{-6}\times\sqrt{12}\)
Evaluate:
\(i^{4}\)
\(i^{9}\)
\(i^{7}-i^{11}\)
\(i^{5}+i^{6}-i^{7}\)
\(2i-i^{6}+2i^{7}\)
State the value of \(\Re(z)\) and \(\Im(z)\) for the following complex numbers.
\(2+7i\)
\(10-i\)
\(\pi+3i\)
\(\dfrac{i}{6}\)
\(-8\)
Find the values of \(x\) and \(y\).
\(x+yi=4+9i\)
\(x+yi=3-i\)
\(x+yi=23\)
\(x+yi=-\sqrt{2}i\)
\(x+i=-5+yi\)
\(3i\)
\(\sqrt{2}i\)
\(\sqrt{15}i\)
\(5\sqrt{2}i\)
\(6\sqrt{2}i\)
\(1\)
\(i\)
\(0\)
\(2i-1\)
\(1\)
\(\Re(z)=2\) and \(\Im(z)=7\)
\(\Re(z)=10\) and \(\Im(z)=-1\)
\(\Re(z)=\pi\) and \(\Im(z)=3\)
\(\Re(z)=0\) and \(\Im(z)=\dfrac{1}{6}\)
\(\Re(z)=-8\) and \(\Im(z)=0\)
\(x=4\) and \(y=9\)
\(x=3\) and \(y=-1\)
\(x=23\) and \(y=0\)
\(x=0\) and \(y=-\sqrt{2}\)
\(x=-5\) and \(y=1\)
Adding and subtracting complex numbers
To add or subtract complex numbers, we add or subtract the real and imaginary parts separately.
Find the conjugate of the following complex numbers.
\(4+9i\)
\(-3-15i\)
\(\sqrt{3}-4i\)
Find the conjugate of \((2-i)(4+7i)\).
If \(z=2-i\) and \(w=1+2i\), express the following in the form \(x+yi\).
\(\overline{z}\)
\(\overline{z+w}\)
\(\overline{z}+\overline{w}\)
\(\overline{zw}\)
\(\overline{\overline{z}-\overline{w}}\)
\(4-9i\)
\(-3+15i\)
\(\sqrt{3}+4i\)
\(15-13i\)
\(2+i\)
\(3-i\)
\(3-i\)
\(4-3i\)
\(1-3i\)
Dividing complex numbers
To divide complex numbers, we rewrite them as fraction and rationalise the same way we do with surds. We multiply both the numerator and denominator by the complex conjugate of the denominator, then simply to the form \(x+yi\).
\[\frac{a+bi}{c+di}\times\frac{c-di}{c-di}\]
Example 1 – dividing complex numbers
Express \(\dfrac{2-i}{1+3i}\) in the form \(x+yi\).
\[\begin{align*} \frac{2-i}{1+3i}\times\frac{1-3i}{1-3i}\\
& = \frac{(2-i)(1-3i)}{(1+3i)(1-3i)}\\
& = \frac{2-6i-i+3i^{2}}{1+9}\\
& = \frac{2-7i-3}{10}\\
& = \frac{-1-7i}{10}\\
& = -\frac{1}{10}-\frac{7}{10}i
\end{align*}\]
Express \(\dfrac{i}{1-4i}\) in the form \(x+yi\).
\[\begin{align*} \frac{i}{1-4i}\times\frac{1+4i}{1+4i}\\
& = \frac{i(1+4i)}{(1-4i)(1+4i)}\\
& = \frac{i+4i^{2}}{1+16}\\
& = \frac{i-4}{17}\\
& = -\frac{4}{17}+\frac{1}{17}i
\end{align*}\]
Exercise – dividing complex numbers
Express the following in the form \(x+yi\).
\(\dfrac{4-9i}{3}\)
\(\dfrac{1}{3-i}\)
\(\dfrac{5+i}{2-7i}\)
Simplify \(\dfrac{2}{1-i}+\dfrac{3+i}{i}\).
If \(w=-1+6i\), express \(\dfrac{w+1}{w-i}\) in the form \(x+yi\).
\(\dfrac{4}{3}-3i\)
\(\dfrac{3}{10}+\dfrac{1}{10}i\)
\(\dfrac{3}{53}+\dfrac{37}{53}i\)
\(2-2i\)
\(\dfrac{15}{13}-\dfrac{3}{13}i\)
Argand diagrams
You can present complex numbers graphically using Argand diagrams. For the complex number \(z=ax+byi\), the \(x\) coordinate will be \(a\) and the \(y\) coordinate will be \(bi\). In other words, the point is \((a,bi)\).
Exercise – drawing Argand diagrams
If \(z=2-3i\) and \(w=1+4i\), illustrate the following on an Argand diagram.