# Imaginary unit

square root of negative one, used to define extra-dimensional complex numbers

In math, imaginary unit or $i$ , is a number value that only exists outside of real numbers and is used in algebra. Though imaginary numbers can be used to solve a lot of mathematical problems, they cannot be represented by an amount of real life objects.

## History

Imaginary units were invented to answer the polynomial equation $x^{2}+1=0$ , which normally has no solution (see below). The term "imaginary" comes from by René Descartes and was meant to be insulting as, like zero and negative numbers at other times in history, imaginary numbers were thought to be useless as they are not natural. It wasn't until later centuries that the work of mathematicians like Leonhard Euler, Augustin-Louis Cauchy and Carl Friedrich Gauss would prove that imaginary numbers were very important for some areas of algebra.

## Definition

A common rule for multiplying and dividing numbers is that if the signs are different then the result is negative (e.g. $4\times -3=-12$ ), but if both numbers have the same sign then the result will be positive (e.g. $5\times 6=30$  and $-10\times -10=100$ ). However, this leads to problems with square root numbers of negatives, as two negative numbers will always make a positive number:

$2\times 2=2^{2}=4$
so ${\sqrt {4}}=2$
but ${\sqrt {-4}}\neq -2$
as $-2\times -2=-2^{2}=4$

To fill in this value gap the imaginary unit was made, which is defined as $i={\sqrt {-1}}$  and $i\times i=i^{2}=-1$ . Using imaginary numbers we can solve our last example:

$2i\times 2i=2i^{2}=-4$
$-2i\times -2i=-2i^{2}=-4$
${\sqrt {4}}=\pm 2$  and ${\sqrt {-4}}=\pm 2i$

## Square root of i

It is sometimes assumed that one must create another number to show the square roots of $i$ , but that is not needed. The square roots of $i$  can be written as: $\pm {\sqrt {i}}=\pm {\frac {\sqrt {2}}{2}}(1+i)$ , a result which can be shown as follows:

 $\left(\pm {\frac {\sqrt {2}}{2}}(1+i)\right)^{2}\$ $=\left(\pm {\frac {\sqrt {2}}{2}}\right)^{2}(1+i)^{2}\$ $=(\pm 1)^{2}{\frac {2}{4}}(1+i)(1+i)\$ $=1\times {\frac {1}{2}}(1+2i+i^{2})\quad \quad (i^{2}=-1)\$ $={\frac {1}{2}}(2i)\$ $=i\$ ## Powers of i

The powers of $i$  or $i$  follow a regular and predictable pattern:

$i^{-4}=1$
$i^{-3}=i$
$i^{-2}=-1$
$i^{-1}=-i$
$i^{0}=1$
$i^{1}=i$
$i^{2}=-1$
$i^{3}=-i$
$i^{4}=1$
$i^{5}=i$
$i^{6}=-1$
$i^{7}=-i$

As shown, each time we multiply by another $i$  the values are $1,i,-1,-i$  and then repeat.