Complex numbers: Construction of complex numbers and complex arithmetic
Conjugate, modulus, and division of complex numbers
Complex numbers are often denoted by the letter \(z\), so \(z=x+y\,\mathrm{i}\).
Conjugate The complex conjugate of \(z=x+y\,\mathrm{i}\), sometimes in short called conjugate of \(z\) and commonly denoted with \(\overline{z}\) or \(z^{\ast}\), is defined as \(\overline{z}=x-y\,\mathrm{i}\).
With certain combinations \(z\) and \(\overline{z}\) something specialis going on and this you might already have noticed in the exercises of addition and multiplication of complex numbers; if not, we give some extra examples.
So \(\frac{1}{2}(z+\overline{z})=\frac{1}{2}\!(1-5\,\mathrm{i}+1+5\,\mathrm{i})=1\)
When you have seen enough examples of the penny dropped: the sum and the product of a complex number with its conjugate is a real number. In fact, for the product the following holds:
For any complex number \(z\) holds \[z\cdot \overline{z}=\bigl(\mathrm{Re}(z)\bigr)^2+\bigl(\mathrm{Im}(z)\bigr)^2\]
If \(z=x+y\,\mathrm{i}\) is a complex number in standard form, then: \[\begin{aligned} z\cdot \overline{z} &= (x+y\,\mathrm{i})\cdot (x-y\,\mathrm{i}) & \color{blue}{\text{definition of conjugate}}\\
&= x^2 - (y\,\mathrm{i})^2 & \color{blue}{\text{special product}}\\
&= x^2-y^2\mathrm{i}^2 & \color{blue}{\text{expansion}}\\ &= x^2+y^2 & \color{blue}{\text{property i}^2=-1}\\
&= \bigl(\mathrm{Re}(z)\bigr)^2+\bigl(\mathrm{Im}(z)\bigr)^2 & \color{blue}{\text{definitions }x=\mathrm{Re}(z)\mathrm{\;and\;}y=\mathrm{Im}(z)}\end{aligned}\]
Modulus
For any complex number \(z\) we call the root \(z\cdot \overline{z}\) the modulus or absolute value of that number and denote this as \(|z|\). For any complex number \(z=x+y\,\mathrm{i}\) holds: \(|z|=\sqrt{x^2+y^2}\).
\(\phantom{x}\)
Now that we know the conjugate and modulus of a complex number, we can easily introduce division of complex numbers. First a simple example of writing \(\dfrac{1}{\mathrm{i}}\) in standard form. The trick is to multiply the numerator and denominator by \(-\mathrm{i}\), the conjugate of the denominator, and to use the rule \(\mathrm{i}^2=-1\)in order to simplify the result:
\[\frac{1}{\mathrm{i}}=\frac{1}{\mathrm{i}}\cdot\frac{-\mathrm{i}}{-\mathrm{i}}=\frac{-\mathrm{i}}{-\mathrm{i}^2}=\frac{-\mathrm{i}}{-(-1)}=-\mathrm{i}\]
For any complex number \(z\) we act similarly, namely multiplying the numerator and denominator by the conjugate of the denominator.
\[\frac{1}{z}=\frac{\overline{z}}{z\cdot \overline{z}}=\frac{ \overline{z}}{|z|^2}\] In other words, the reciprocal value of a complex number is a real multiple of the conjugate. Totally written out we get:
If \(z=x+y\,\mathrm{i}\). then \[\begin{aligned}\frac{1}{z}&=\frac{ \overline{z}}{z\cdot \overline{z}} \\ \\ &=\frac{\overline{z}}{|z|^2} \\ \\ &= \frac{x-y\,\mathrm{i}}{x^2+y^2}\\ \\ &= \frac{x}{x^2+y^2}- \frac{y}{x^2+y^2}\cdot\mathrm{i}\end{aligned}\]
For two complex numbers \(w\) and \(z\) we perform division as follows: \[\frac{w}{z}=\frac{w\cdot \overline{z}}{z\cdot \overline{z}}=\frac{1}{|z|^2}\cdot w\cdot\overline{z}\]
Fully written out, division of two complex numbers goes as follows:
Division of complex numbers as \(w=u+v\,\mathrm{i}\) and \(z=x+y\,\mathrm{i}\) are complex numbers in standard form, then division goes as follows: \[\begin{aligned}\frac{w}{z} &= \frac{u+v\,\mathrm{i}}{x+y\,\mathrm{i}} \\ \\ &= \frac{(u+v\,\mathrm{i})\cdot (x-y\,\mathrm{i})}{(x+y\,\mathrm{i})\cdot (x-y\,\mathrm{i})}\\ \\ &= \frac{u\cdot x+v\cdot y + (x\cdot v-u\cdot y)\,\mathrm{i}}{x^2+y^2}\\ \\ &=\frac{u x+v y}{x^2+y^2} + \frac{x v-u y}{x^2+y^2}\mathrm{i}\end{aligned}\]
You do not need to memorize this formula; it suffices to know the concept and be able to apply the method. In fact, it suffices in division of two complex numbers to multiply the numerator and denominator by the complex conjugate of the denominator, then to exapand brackets and make use of the rule that \(\mathrm{i}^2=-1\). Some sample calculations illustrate this.
\[\begin{aligned}
\frac{1}{4-3\,\mathrm{i}}&= \frac{1}{4-3\,\mathrm{i}}\cdot \frac{4+3\,\mathrm{i}}{4+3\,\mathrm{i}} &\color{blue}{\text{use of conjugate}}\\ \\
&= \frac{4+3\,\mathrm{i}}{(4-3\,\mathrm{i})\cdot(4+3\,\mathrm{i})} &\color{blue}{\text{multiplication of fractions}}\\ \\
&= \frac{4+3\,\mathrm{i}}{4^2-(3\,\mathrm{i})^2}&\color{blue}{\text{special product in denominator}}\\ \\
&= \frac{4+3\,\mathrm{i}}{4^2-3^2\cdot \mathrm{i}^2}&\color{blue}{\text{expansion of denominator}}\\ \\
&=\frac{4+3\,\mathrm{i}}{4^2+3^2}&\color{blue}{\text{property i}^2=-1}\\ \\
&=\frac{4+3\,\mathrm{i}}{16+9}&\color{blue}{\text{expansion of denominator}}\\ \\
&=\frac{4+3\,\mathrm{i}}{25}& \color{blue}{\text{simplification of denominator}}\\ \\
&=\frac{4}{25}+\frac{3}{25}\,\mathrm{i} & \color{blue}{\text{rewriting into standard form}}
\end{aligned}\]
Mathcentre videos
Complex Conjugate (5:23)
Division (7:23)
