Home Algebra Introduction to Complex numbers

# Introduction to Complex numbers We propose an introduction to complex numbers. Such numbers allow finding a bridge with geometry, especially in the field of rotations and similarities. This is a generalization of real numbers.

In principle, the complex plan allows the resolution of problems practically insoluble otherwise. It is the case of solving equations of the form $x^2+1=0$, for example. Or some times it is very hard to calculate certain integrals, then a part of the complex analysis theory was created to calculate such integrals.

## An introduction to complex numbers

A complex number is an element of the form $z=a+ib$ where $a$ and $b$ are real numbers and $i$ is a a typical element satisfying $i^2=-1$. The set of all complex numbers will be denoted by $\mathbb{C}$. The number $a$ is called the real part of the complex number $z$ and will be denoted by ${\rm Re}(z),$ while the real number $b$ is called the imaginary part of $z$ and will be denoted by ${\rm Im}(z)$.

On complex numbers set $\mathbb{C},$ we define to operation “$+$” and “$\cdot$” as follows: if $z_1=a_1+ib_1,$ $z_2=a_2+ib_2$ and $\lambda\in\mathbb{R},$ \begin{align*} z_1+z_2=(a_1+a_2)+i(b_1+b_2),\quad \lambda\cdot z_1=(\lambda a_1)+(\lambda b_1).\end{align*} We have $(\mathbb{C},+,\cdot)$ is a vector space on the field $\mathbb{R}$.

We then constructed a natural linear map $f:\mathbb{R}\times \mathbb{R}\to \mathbb{C}$ by selecting $f(a,b)=a+ib$. By construction $f$ is surjective. Let us verify that $f$ is injective. As $f$ is a linear map, we determine its kernel. If $f(a,b)=0,$ then $a+ib=0$. By multiplying both sides of this equation by $i,$ we get  $-b+ia=0$. By adding the two equations, we obtain $(a-b)(1-i)=0$. As $1-i\neq 0$, then $a-b=0$.

We conclude that the complex set $\mathbb{C}$ is in bijection with the plane $\mathbb{R^2}$. Thus $\mathbb{C}$ is the plane. Thus any complex number $z=a+ib$ corresponds to a point $M$ in the plane with coordinates $a$ and $b$, we write $M(a,b)$.

## Operations on complex numbers

To multiply two complex numbers $z_1=a_1+ib_1,$ $z_2=a_2+ib_2$, we use the same calculus disctribution as in $\mathbb{R}$ and we use the fact that $i^2=-1,$ we obtain \begin{align*} z_1 z_2=(a_1a_2-b_1b_2)+i(a_1b_2+a_2b_1).\end{align*} In particular, we define de square of the complex number $z=a+ib$ by \begin{align*} z^2=z z= (a^2-b^2)+i (2ab).\end{align*}

### How to determine the Square root of a complex number

Let us work on a simple example: assume we want to define a calculate the square root of a complex number $\lambda=4+3i$. For that purpose, you should look for a complex number $z$ such that $z^2=\lambda$. If we select $z=a+ib$ with $a$ and $b$ are real numbers, then we need to determine $a$ and $b$. Observe that $z^2=a^2+2iab+(ib)^2$. As $i^2=-1,$ we have $z^2=a^2-b^2+i(2ab)$.

On the other hand, let us recall that $s+it=s’+it’$ for real numbers $s,t,s’,t’$, then $s=s’$ and $t=t’$. Using this identification, the fact that $a^2-b^2+i(2ab) =4+i3$ implies that\begin{align*} a^2-b^2=4\quad\text{and}\quad 2ab=3.\end{align*} Now we give the key of solving the above system. We need a third equation involving the $a^2$ and $b^2$. We obtain this equation by taking the modal of $z$, this is $|z|^2=a^2+b^2=| 4+i3 |$. Hence \begin{align*}a^2+b^2=\sqrt{16+9}=5.\end{align*}

The role of this equation is to eliminate one of the squares $a^2$ or $b^2$. In fact, by adding the two sides of the equations containing $a^2$ and $b^2,$ we obtain $2a^2=9$, so that $a=\pm \frac{3}{2}\sqrt{2}$. Moreover, $2b^2=2a^2-4=9-4=5$. This shows that $\pm \sqrt{\frac{5}{2}}$. Finally, the number $\lambda$ has two squat root given by\begin{align*} \frac{3}{2}\sqrt{2} +i \sqrt{\frac{5}{2}},\quad – \frac{3}{2}\sqrt{2} -i \sqrt{\frac{5}{2}}.\end{align*}

### How to solve a system of complex numbers

To conclude this introduction to complex numbers, we will learn you how to solve a system of complex equations. As a matter of facts, systems of equations with real numbers are standrad. One can use determinants or elemination technique to solve such systems. In the sequel we are interested in solving systems with the category of complex numbers. To simplify the task we shall work on a particuar system. Le consider the system\begin{align*}z_1z_2=i,\qquad z_1-z_2=1+i,\end{align*} where $z_1$ and $z_2$ are complex numbers. Let us determine the expressions of $z_1$ and $z_2$.

In this question, we must be careful and not go into hard calculations. In fact, it would be awkward to determine $z_1$ as a function of $z_2,$ and then replace the expression of $z_1$ in the second equation. We will introduce a nice and short method to solve the complex system. We have\begin{align*} z_1 (-z_2)=-i,\quad z_1+(-z_2)=1+i.\end{align*} We deduce that $z_1$ et $(-z_2)$ are solutions of the equation:

\begin{align*}\tag{E}t^2-(1+i)t-i=0.\end{align*} The discriminant associated with this equation is\begin{align*} \Delta&= 2i+4i=6i\cr &= \left(\sqrt{6}\left(\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2} i\right)\right)^2\end{align*} Hence $\Delta$ admits as square roots:\begin{align*}\Delta_1=\sqrt{3}+\sqrt{3}i\quad \text{\nd}\quad \Delta_2=-\sqrt{3}-\sqrt{3}i.\end{align*} The roots of the equation $(E)$ are\begin{align*}z’&=\frac{1+i+\Delta_1}{2}\cr &= \frac{1+\sqrt{3}}{2}(1+i)\end{align*} and \begin{align*}z”&=\frac{1+i+\Delta_2}{2}\cr &= \frac{1-\sqrt{3}}{2}(1+i).\end{align*} Thus, we have $z_1=z’$ and $-z_2=z”$ or $z_1=z”$ and $-z_2=z’$. So the solutions of the system are pair $(z’,-z”)$ and $(z”,-z’),$\begin{align*} &\left(\frac{1+\sqrt{3}}{2}(1+i), \frac{\sqrt{3}-1}{2}(1+i)\right),\;\text{and}\cr
&\left(\frac{1-\sqrt{3}}{2}(1+i),\frac{-1-\sqrt{3}}{2}(1+i)\right). \end{align*}