Complex numbers: senior math course download pdf.

A high school math course on complex numbers.

This lesson involves the following concepts:

  • definition of the complex number;
  • algebraic form;
  • geometric form;
  • Euler’s formula;
  • Moivre’s formula;
  • complex equations;
  • geometric representation of a complex number;
  • real and imaginary part of a complex number;
  • operations on complex numbers.

I. Algebraic form of a complex number

Theorem and definition:

There exists a set of numbers noted \mathbb{C}, whose elements are called the complex numbers, such that :

  • \mathbb{C} contains the set \mathbb{R} of real numbers;
  • the calculation rules in \mathbb{C} are the same as in \mathbb{R};
  • \mathbb{C} contains an element noted i such that i^2=-1;
  • any complex number z can be written as in a unique way in the form z=x+iy with x and y The number x is called the real part (noted Re(z)) of the number z and the number y is called the imaginary part (noted Im(z)) of the complex number z.

sets of numbers


The number z=\sqrt{3}+2i is a complex number.

\sqrt{3} is its real part and 2 is its imaginary part.

  • z is a real number if and only if Im(z)=0.
  • z is a pure imaginary if and only if Re(z)=0.

II. conjugate of a complex number


We consider z a complex number whose algebraic form is z=x+iy with x and y two real numbers.We call conjugate of the number z, the complex number, noted \overline{z}, such as \overline{z}=x-iy.


\overline{1+3i}=1-3i and \overline{2-5i}=2+5i.


We consider two complex numbers z and z'. We have the following properties:

  • \overline{\overline{z}}=z
  • \overline{z+z'}=\overline{z}+\overline{z'}
  • \overline{(\frac{1}{z})}=\frac{1}{\overline{z}} with z\neq\,0
  • z\in\,\mathbb{R}\Leftrightarrow\,\overline{z}=z
  • z is a pure imaginary \Leftrightarrow\,\overline{z}=-z
  • \overline{zz'}=\overline{z}\overline{z'}
  • \overline{(\frac{z}{z'})}=\frac{\overline{z}}{\overline{z'}} with z'\neq\,0
  • \overline{(z^n)}=(\overline{z})\,^n with n\in\,\mathbb{N}
  • \overline{(kz)}=k\,\overline{z} with k\in\,\mathbb{R}

III.graphical representation of complex numbers

1. Affix of a point


Consider the complex plane with a direct orthonormal reference frame (O,\vec{u},\vec{v})

To any complex number z=x+iy , we associate the point M(x;y).

M is called the image point of z and z is called theaffix of the point M in the direct orthonormal reference frame (O,\vec{u},\vec{v}). Let M(z) be the point M with affix z.

Graphical representation of complex numbers


The point M with affix z=3+i has coordinates M(3,1).

The point N with affix z=-1-i has coordinates M(-1,-1).

2. affix of a vector


To any complex number z affix of the point M(x,y), we associate the vector \vec{w}=\vec{OM} such that \vec{w}(x;y).and we note \vec{\,w}(z), the vector \vec{\,w} of affix z.


The vector \vec{OM} with affix z=1+2i has coordinates \vec{\,OM}(1;2).

The vector \vec{t} with affix 1-3i has coordinates \vec{\,t}(1;-3).

vector affix


We consider two vectors \vec{w} and \vec{w'} of respective affixes z andz'The vector \vec{w}+\vec{w'} has the affix z+z'.

The vector k\vec{w} has affix kz with k\in\,\mathbb{R}.

3. second degree equations in \mathbb{C}


Consider a real number a.

  • If a>0, the solutions are z=\sqrt{a} and z=-\sqrt{a};
  • If a<0, the solutions are z=i\sqrt{-a} and z=-\sqrt{-a};
  • If a=0, the solution is z=0.


The equation z^2=-4 has solutions in \mathbb{C}: z=2i and z=-2i.

4. equations of the type az²+bz+c=0


We consider real numbers a, b and c with a\neq\,0. We consider in \mathbb{C}, the equation (E) : az^2+bz+c=0 of discriminant \Delta\,=b^2-4ac.

  • If \Delta>0, the solutions are z_1=\frac{-b+\sqrt{\,\Delta\,}}{2a} and z_2=\frac{-b-\sqrt{\,\Delta\,}}{2a};
  • If \Delta<0, the solutions are z_1=\frac{-b+i\sqrt{-\,\Delta\,}}{2a} and z_2=\frac{-b-i\sqrt{\,-\Delta\,}}{2a};
  • If \Delta=0, the solution is z\,=\frac{\,\sqrt{\,\Delta\,}}{2a}.


Solve in \mathbb{C}, the equation (E) : z^2+4z+5=0.

\Delta\,=b^2-4ac=4^2-4\times  \,1\times  \,5=16-20=-4<0.

The solutions are:


and .

5.factoring of a second degree trinomial


We consider real numbers a,b and c with a\neq\,0. For any number z\in\,\mathbb{C}, we pose P(z)=az^2+bz+c.

We note z_1 and z_2 the two solutions of P(z)=0 in (with possibly z_1= z_2 when \Delta=0).

We have for all z\in\,\mathbb{C}, P(z)=a(z-z_1)(z-z_2).


Let’s go back to the previous example, P(z)=z^2+4z+5=\,(z+2-i)(z+2+i).

Cette publication est également disponible en : Français (French) العربية (Arabic)

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