# Imaginary Numbers: Properties, Applications, Examples

The **imaginary numbers**are those that give a solution to the equation in which the unknown, squared, is equal to a negative real number. The imaginary unit is*i = √ (-1)*.

In the equation: *z ^{2} = – a, *

*z*is an imaginary number that is expressed as follows:

*z = √ (-a) = i√ (a)*

Being *to*a positive real number. Yes*a = 1*, so *z = i*, where *i* is the imaginary unit.

In general, a pure imaginary number z is always expressed in the form:

*z = y⋅i*

Where *and* is a real number e *i* is the imaginary unit.

Just as real numbers are represented on a line, called the *real straight*, in an analogous way the imaginary numbers are represented on the *imaginary line*.

The *imaginary line* is always orthogonal (90º shape) to the *real straight* and the two lines define a Cartesian plane called the *complex plane*.

In figure 1 the complex plane is shown and on it some real numbers, some imaginary numbers and also some complex numbers are represented:

X _{1} , X _{2} , X _{3} are real numbers

Y _{1} , Y _{2} , Y _{3} are imaginary numbers

Z _{2} and Z _{3} are complex numbers

The number O is the real zero and it is also the imaginary zero, so the origin O is the complex zero expressed by:

0 + 0i

**Properties**

The set of imaginary numbers is denoted by:

I = {……, -3i,…, -2i,…., – i,…., 0i,…., I,…., 2i,…., 3i, ……}

And you can define some operations on this numerical set. An imaginary number is not always obtained from these operations, so let’s look at them in a little more detail:

**Add and subtract imaginary**

Imaginary numbers can be added and subtracted from each other, resulting in a new imaginary number. For example:

3i + 2i = 5i

4i – 7i = -3i

**Product of imaginary**

When the product of one imaginary number with another is made, the result is a real number. Let’s do the following operation to check it:

2i x 3i = 6 xi ^{2} = 6 x (√ (-1)) ^{2} = 6 x (-1) = -6.

And as we can see, -6 is a real number, although it has been obtained by multiplying two pure imaginary numbers.

**Product of a real number by another imaginary**

If a real number is multiplied by i, the result will be an imaginary number, which corresponds to a 90-degree counterclockwise rotation.

And it is that i ^{2} corresponds to two consecutive rotations of 90 degrees, which is equivalent to multiplying by -1, that is, i ^{2} = -1. It can be seen in the following diagram:

For example:

-3 x 5i = -15i

-3 xi = -3i.

**Empowerment of an imaginary**

You can define the potentiation of an imaginary number to an integer exponent:

*i ^{1} = i*

*i ^{2} = ixi = √ (-1) x √ (-1) = -1*

*i ^{3} = ixi ^{2} = -i*

*i ^{4} = i ^{2} xi ^{2} = -1 x -1 = 1*

*i ^{5} = ixi ^{4} = i*

In general you have to *i ^{n} = i ^ (n mod 4),* where

*mod*is the remainder of the division between

*n*and

*4*.

Negative integer potentiation can also be done:

i ^{-1} = 1 / i ^{1} = i / (ixi ^{1} ) = i / (i ^{2} ) = i / (-1) = -i

i- ^{2} = 1 / i ^{2} = 1 / (-1) = -1

i- ^{3} = 1 / i ^{3} = 1 / (- i) = (-1) / i = -1 xi ^{-1} = (-1) x (-i) = i

In general, the imaginary number b⋅i raised to the power n is:

(b⋅i) i ^{n} = b ^{n} i ^{n} = b ^{n} i ^ (n mod 4)

Some examples are the following:

(5 i) ^{12} = 5 ^{12} i ^{12} = 5 ^{12} i ^{0} = 5 ^{12} x 1 = 244140625

(5 i) ^{11} = 5 ^{11} i ^{11} = 5 ^{11} i ^{3} = 5 ^{11} x (-i) = -48828125 i

(-2 i) ^{10} = -2 ^{10} i ^{10} = 2 ^{10} i ^{2} = 1024 x (-1) = -1024

**Sum of a real number and an imaginary number**

When you add a real number with an imaginary one, the result is neither real nor imaginary, it is a new type of number called a *complex number* .

For example, if X = 3.5 and Y = 3.75i, then the result is the complex number:

Z = X + Y = 3.5 + 3.75 i

Note that in the sum the real and imaginary parts cannot be grouped together, so a complex number will always have a real part and an imaginary part.

This operation extends the set of real numbers to the widest of complex numbers.

**Applications**

The name of imaginary numbers was proposed by the French mathematician René Descartes (1596-1650) as a mockery or disagreement with the proposal of the same made by the Italian mathematician of the century Raffaelle Bombelli.

Other great mathematicians, such as Euler and Leibniz, supported Descartes in this disagreement and called imaginary *numbers amphibious numbers, * which were torn between being and nothing.

The name of imaginary numbers remains today, but their existence and importance is very real and palpable, since they appear naturally in many fields of physics such as:

-The theory of relativity.

-In electromagnetism.

-Quantum mechanics.

**Exercises with imaginary numbers**

**– Exercise 1**

Find the solutions of the following equation:

z ^{2} + 16 = 0

**Solution**

z ^{2} = -16

Taking square root in both members we have:

√ (z ^{2} ) = √ (-16)

± z = √ (-1 x 16) = √ (-1) √ (16) = ix 4 = 4i

In other words, the solutions of the original equation are:

z = + 4i oz = -4i.

**– Exercise 2**

Find the result of raising the imaginary unit to the power 5 minus the subtraction of the imaginary unit raised to the power -5.

**Solution**

i ^{5} – i- ^{5} = i ^{5} – 1 / i ^{5} = i – 1 / i = i – (i) / (ixi) = i – i / (- 1) = i + i = 2i

**– Exercise 3**

Find the result of the following operation:

(3i) ^{3} + 9i

**Solution**

3 ^{3} i ^{3} – 9 = 9 (-i) + 9i = -9i + 9i = 0i

**– Exercise 4**

Find the solutions of the following quadratic equation:

(-2x) ^{2} + 2 = 0

**Solution**

The equation is rearranged as follows:

(-2x) ^{2} = -2

Then the square root of both members is taken

√ ((- 2x) ^{2} ) = √ (-2)

± (-2x) = √ (-1 x 2) = √ (-1) √ (2) = i √ (2) = √2 i

Then we solve for x to finally obtain:

x = ± √2 / 2 i

That is, there are two possible solutions:

x = (√2 / 2) i

Or this other:

x = – (√2 / 2) i

**– Exercise 5**

Find the value of Z defined by:

Z = √ (-9) √ (-4) + 7

**Solution**

We know that the square root of a negative real number is an imaginary number, for example √ (-9) is equal to √ (9) x √ (-1) = 3i.

On the other hand, √ (-4) is equal to √ (4) x √ (-1) = 2i.

So the original equation can be replaced by:

3i x 2i – 7 = 6 i ^{2} – 7 = 6 (-1) – 7 = -6 – 7 = -13

**– Exercise 6**

Find the value of Z resulting from the following division of two complex numbers:

Z = (9 – i^{two}) / (3 + i)

**Solution**

The numerator of the expression can be factored using the following property:

*A difference of squares is the product of the sum and the difference of the binomials without squaring.*

So:

Z = [(3 – i) (3 + i)] / (3 + i)

The resulting expression is simplified below, leaving

Z = (3 – i)

**References**

- Earl, R. Complex numbers. Recovered from: maths.ox.ac.uk.
- Figuera, J. 2000. Mathematics 1st. Diversified. CO-BO editions.
- Hoffmann, J. 2005. Selection of Mathematics topics. Monfort Publications.
- Jiménez, R. 2008. Algebra. Prentice Hall.
- Wikipedia. Imaginary number. Recovered from: en.wikipedia.org