# Definition of complex numbers

We know that $2\cdot 2 = 4$, so $2$ is the square root of $4$. We know that $5\cdot 5 = 25$, so $5$ is the square root of $25$.

But, what happens when we try to determine the square root of $-4$? How could we find a number that when multiplied by itself gives $-4$?

As seen in square root subject, there doesn't exist any real number solution of a negative square root. That is, square roots like $\sqrt{-4}$ or $\sqrt{-25}$ don't have any solutions in the set of real numbers. This is why imaginary numbers (or complex numbers) were invented, in order to give solutions to square roots of negative numbers. Any negative number's square root can be written as $\sqrt{-a}$ and, using square root's properties, it can be expressed:

$$\sqrt{-a}=\sqrt{(a)\cdot(-1)}=\sqrt{a}\cdot\sqrt{-1}$$

As we already know to calculate positive numbers' square roots, the only thing left was to invent the square root of $-1$. This number was called imaginary unit and it is written $i$.

$$\sqrt{-a}=\sqrt{a}\cdot\sqrt{-1}=\sqrt{a}\cdot i\quad \text{given that} \quad i=\sqrt{-1}$$

The complex numbers, also known as imaginary numbers, serve to get the square root of negative numbers. Thanks to them we can find solutions to functions that we previously did not know how to solve because they did not have a real solution.

For example, to find the solution to the equation:

$$x^2+1=0 \Rightarrow x^2 = -1 \Rightarrow x=\sqrt{-1}$$

This is how we define the number $i$, known as the imaginary unit, which is $i=\sqrt{-1}$ which allows us to say that the previous equation of the second degree has a solution, or also, it allows us to compute the square root of any negative number.

Thus we have, for example:

1. $\sqrt{-144}=\sqrt{144}\cdot\sqrt{-1}=12i$
2. $x^2+64=64$ has solution since $x^2=-64 \Rightarrow x=\sqrt{-64} =\sqrt{(-1)\cdot64} \Rightarrow x=8\cdot\sqrt{-1}= 8i$

Now that we know what the imaginary unit is, we can introduce all the complex numbers.

A complex number is described as the sum of a real number and an imaginary number (that is, a multiple of the imaginary unit, which is indicated by the letter $i$). Therefore, the complex numbers are represnted by $z$ and are composed of:

$$z=a+bi$$

where we call:

• $a$ is the real part.
• $b$ is the imaginary part (that is, the coefficient multiplying the imaginary unit $i$).

Let's see some examples of imaginary numbers expressed in this form $z=a+bi$ which we will call binomial form.

$3+6i$ is the complex number with real part $3$ and imaginary part $6$.

$-5+\dfrac{\sqrt{5}}{23}i$ is the complex number with real part $-5$ and imaginary part $\dfrac{\sqrt{5}}{23}$.

Let's see some special cases.

If $b = 0$, the complex number is of the kind $z=a+0\cdot i=a$ therefore, it is a real number, since the imaginary part does not exist. That is why we say that the real numbers are a subset of the complex numbers.

$2$, $-7$, $\sqrt{5}$, $\dfrac{3}{2}$ are real numbers, and therefore, complexes with the imaginary part zero.

If $a = 0$, the complex number is of the kind $z=0+b\cdot i=bi$, therefore, it is a multiple of the imaginary unit, and it is said that it is a pure imaginary number.

$2i$, $-7i$, $\sqrt{-5}$, $\dfrac{3i}{2}$ are pure imaginary numbers because they are multiples of the imaginary unit $i$.

If $a = b = 0$, we have the complex number $z=0+0\cdot i=0$, which is called the complex number zero.

The complex or imaginary numbers are an extension of the real numbers, characterized by the fact that they give all the roots of the polynomials. This is to say, for any polynomial with real coefficients, it will always have all the solutions in the set of complex numbers. Note that in the complex numbers we do not have a total order, as we do with real numbers. This means that when we have two real numbers we can always tell which is the greatest. This is no longer true with complex numbers. We can establish, however, a criterion to know if two complex numbers are the same or not:

• The real parts of two numbers must be identically equal.
• The imaginary parts of two numbers must be also identically equal.

Namely:

$$a+bi=a'+b'i \ \Longleftrightarrow \ a=a' \ \text{and} \ \ b=b'$$

$3+5i=\dfrac{9}{3} + 5i$ they are equal since $3=\dfrac{9}{3}$ and $5=5$.

$2-8i\neq 2+8i$ they are not equal because the imaginary factor of the first one is $-8$ and that of the second one is $+8$.

$\dfrac{55+11i}{11}=5-(-i)$ they are equal since $\dfrac{55}{11}=5$ and $\dfrac{11i}{11}=-(-i)=i$