A complex number is any number that can be written in the form $z = a + b i$ where $a$ is called the 'real' part and $b i$ the 'imaginary' part. The complex number $z = a + b i$ is said to be written in real-imaginary form.
The imaginary unit $i$ is defined to be $i=\sqrt{-1}$. The imaginary unit is sometimes labelled $\mathrm{j}$, usually in engineering.
A complex number whose real part is zero is said to be purely imaginary, whereas a complex number whose imaginary part is zero is a real number. A number of the form $z = a + b i$
The set of all complex numbers is denoted by $\C$.
Note: The terms 'real' and 'imaginary' are artefacts of historical objections to the use of complex numbers, which arose when the interpretation of negative square roots was not well understood. 'Imaginary' numbers are no less real than 'real' numbers!
The complex conjugate of a complex number $z$ is denoted by $\bar{z}$ or $z^{\ast}$. If $z = a + bi$ then its complex conjugate is \[\bar{z} = a - bi\]
Some properties of conjugation are:
\begin{align} \bar{\bar z} &= z \\ \\ z \cdot \bar{z} &= a^2+b^2 \\ \\ \overline{z+w} &= \bar{z} + \bar{w} \\ \\ \overline{zw} &= \bar{z} \cdot \bar{w} \\ \\ \overline{\left(\dfrac{z}{w}\right)} &= \dfrac{\bar{z} }{\bar{w} } \end{align}
Given two complex numbers $z_1 = a + b i$ and $z_2 = c + d i$, their sum and difference is found by adding or subtracting their real and imaginary parts separately:
\begin{align} z_1 + z_2 &= (a + c) + (b + d) i \\ z_1 - z_2 &= (a - c) + (b - d) i \end{align}
Let $z_1 = 3+2 i$ and $z_2 = -4+ i$, then work out:
$z_1 + z_2$;
\begin{align} z_1 + z_2 &= (3 + 2 i) + (-4+ i)\\ &=3+2 i-4+ i\\ &=-1 +3 i \end{align}
Let $z_1 = 3+2 i$ and $z_2 = -4+ i$, then work out:
$z_1-z_2$;
\begin{align} z_1-z_2 &= (3+2 i)-(-4+ i)\\ &=3+2 i+4- i\\ &=7+ i \end{align}
Given two complex numbers $z_1 = a + b i$ and $z_2 = c + d i$, the method for complex multiplication is the same as expanding brackets: \begin{align} z_1z_2 &= (a+b i)(c+d i)\\ &= ac + (ad) i + (bc) i + (bd) i^2\\ &= ac + (ad+bc) i - bd & \text{(as }i^2=-1 \text{)}\\ &=(ac-bd)+(ad+bc) i \end{align}
Note: The order of multiplication doesn't matter: $z_1z_2=z_2z_1$.
When multiplying a complex number $z = a + b i$ by a real number $\lambda$, the rule is simpler: \[\lambda z =\lambda a + \lambda b i\]
Let $z = 3-2 i$, then find:
a) $4z$
b) $-2z$
c) $i z$
a) \begin{align} 4z &= 4(3-2i)\\ &= 3\times 4 - 2i \times 4\\ &= 12-8i \end{align} b) \begin{align} -2z &= -2(3 - 2i)\\ &= 3 \times (-2) - 2i \times (-2)\\ &= -6 + 4i \end{align} c) \begin{align} i z &= i(3 - 2i)\\ &= 3 \times i - 2i \times i\\ &= 3i - 2i^2\\ &= 2+3i \end{align}
Let $z_1=3-i$ and $z_2=-5+2i$. Find $z_1z_2$.
\begin{align} z_1z_2 &= (3-i)(-5+2i)\\ &=-15 +6i +5i -2 i^2\\ &=-15 +11i +2\\ &=-13+11i \end{align}
Let $z_1 = 4+2i$ and $z_2 = 4-2i$, then work out $z_1z_2$.
\begin{align} z_1z_2 &= (4+2i)(4-2i)\\ &=16 -8i +8i -4i^2\\ &=20 \end{align}
The inverse, or reciprocal, of a number is the number it must be multiplied by to get $1$.
For example, the inverse of $2$ is $\dfrac{1}{2}$, as $2 \times \dfrac{1}{2} = 1$.
The same can be applied to complex numbers. The inverse of a complex number $z$ is denoted $z^{-1}$.
Let $z=a+bi$, then
\[z^{-1} = \frac{1}{a+bi}\]
It would be convenient to write this in real-imaginary form $c+di$. We will now see how to do that.
Multiply the top and bottom by the complex conjugate of $z$; $a-bi$:
\[\frac{1}{a+bi} \times \frac{a-bi}{a-bi}\] \[\frac{a-bi}{(a+bi)(a-bi)}\]
Expand the brackets.
\[\frac{a-bi}{a^2-abi+abi-b^2i^2}\]
Tidying this up and using the fact that $i^2=-1$, we get
\[z^{-1} = \frac{a-bi}{a^2+b^2}=\left(\frac{a}{a^2+b^2}\right) - \left(\frac{b}{a^2+b^2}\right)i\]
which is now in real-imaginary form.
Find the inverse of the complex number $z=2-3i$.
\[z^{-1} = \frac{1}{2-3i}\]
To write this in real-imaginary form, multiply the top and bottom of the fraction by the complex conjugate $\bar z = 2+3i$.
\begin{align} z^{-1} &= \frac{1}{2-3i}\times \frac{2+3i}{2+3i} \\ &= \frac{2+3i}{(2-3i)(2+3i)} \end{align}
Expand the brackets and tidy up. \begin{align} z^{-1} &=\frac{2+3i}{4 -6i + 6i -9i^2}\\\\ &=\frac{2+3i}{4+9}\\\\ &=\frac{2+3i}{13}\\ &= \frac{2}{13}+\frac{3}{13} i \end{align}
Find the inverse of the complex number $z=-4+5i$.
\[z^{-1} = \frac{1}{-4+5 i}\]
To write this in real-imaginary form, multiply the top and bottom of the fraction by the complex conjugate $\bar z = -4 - 5i$. \begin{align} z^{-1} &= \frac{1}{-4+5i} \times \frac{-4-5i}{-4-5i} \\ &= \frac{-4 - 5i}{(-4 + 5i)(-4 - 5i)} \end{align}
Expand the brackets and tidy up. \begin{align} z^{-1} &= \frac{-4-5i}{16-20i+20i -25i^2}\\\\ &=\frac{-4-5i}{16+25}\\\\ &=\frac{-4-5i}{41}\\ &=-\frac{4}{41} - \frac{5}{41}i \end{align}
Given two complex numbers $z_1 = a + b i$ and $z_2 = c + di$, we can define division as: \[\frac{z_1}{z_2} = \frac{a+bi}{c+di}\]
Again, it would be useful to have this in real-imaginary form.
First multiply both the denominator and numerator of the fraction by the complex conjugate of the denominator, $c-di$. This is done so that the imaginary part in the denominator disappears, as $z_2\bar{z_2}=|z_2|^2$ is a pure real number.
\[\frac{a+bi}{c+di} \times \frac{c-di}{c-di} = \frac{(a+bi)(c-di)}{(c+di)(c-di)}\]
Using the multiplication rules from above, and the fact that $i^2 = -1$, we obtain
\[\frac{(a+bi)(c-di)}{(c+di)(c-di)}=\frac{ac-adi+bci-bdi^2}{c^2-cdi+cdi-d^2i^2}=\frac{(ac + bd) + (bc - ad)i}{c^2+d^2}\]
Now, writing the complex number in real-imaginary form, we obtain \[\frac{z_1}{z_2}= \left( \frac{ac+bd}{c^2+d^2} \right) + \left( \frac{bc-ad}{c^2+d^2} \right) i\]
However it is best to remember the method rather than the formula.
Find the inverse of the complex number $z=2-3i$.
\[z^{-1} = \frac{1}{2-3i}\]
To write this in real-imaginary form, multiply the top and bottom of the fraction by the complex conjugate $\bar z = 2+3i$.
\begin{align} z^{-1} &= \frac{1}{2-3i}\times \frac{2+3i}{2+3i} \\ &= \frac{2+3i}{(2-3i)(2+3i)} \end{align}
Expand the brackets and tidy up. \begin{align} z^{-1} &=\frac{2+3i}{4 -6i + 6i -9i^2}\\\\ &=\frac{2+3i}{4+9}\\\\ &=\frac{2+3i}{13}\\ &= \frac{2}{13}+\frac{3}{13} i \end{align}
Find the inverse of the complex number $z=-4+5i$.
\[z^{-1} = \frac{1}{-4+5 i}\]
To write this in real-imaginary form, multiply the top and bottom of the fraction by the complex conjugate $\bar z = -4 - 5i$. \begin{align} z^{-1} &= \frac{1}{-4+5i} \times \frac{-4-5i}{-4-5i} \\ &= \frac{-4 - 5i}{(-4 + 5i)(-4 - 5i)} \end{align}
Expand the brackets and tidy up. \begin{align} z^{-1} &= \frac{-4-5i}{16-20i+20i -25i^2}\\\\ &=\frac{-4-5i}{16+25}\\\\ &=\frac{-4-5i}{41}\\ &=-\frac{4}{41} - \frac{5}{41}i \end{align}
Prof. Robin Johnson performs complex arithmetic on the complex numbers $2+3i$ and $3-i$.
These workbooks produced by HELM are good revision aids, containing key points for revision and many worked examples.
Test yourself: Numbas test on multiplying and dividing complex numbers