How do you simplify \[{{i}^{5}}?\]

Hint: A complex number is the combination of an imaginary number and the real number.
The given term $i$in the question is the complex number and known. A number in the form of $a+ib$is called a complex number. The term with $i$(iota) is called the imaginary part and the first term $a$is a real part. The complex number is represented by $Z$. And it is expressed as $Z=a+ib$.
The value of $i$is equal to $\sqrt{-1}$
If the imaginary part of a complex number will be zero then the complex number will be a purely real number. And if the real part of the complex number is zero then the complex number will be purely imaginary.
As we know that the value of $i$is equal to $\sqrt{-1}$
$\Rightarrow i=\sqrt{-1}$.

Complete step by step solution:
The given term in the question is ${{i}^{5}}$
To simplify the above term first of all we must have to know the property of $i$that are
We know that $i=\sqrt{-1}$
\[\Rightarrow {{i}^{2}}=i\times i=\sqrt{-1}\times \sqrt{-1}=(-1)\]
\[\Rightarrow {{i}^{3}}={{i}^{2}}\times i=(-1)\times \sqrt{-1}=-\sqrt{-1}=-i\]
\[\Rightarrow {{i}^{4}}={{i}^{2}}\times {{i}^{2}}=(-1)\times (-1)=1\]
Thus to simplify ${{i}^{5}}$first we have to do the factor $5$.
 We can write $5$in the factors form as
$\Rightarrow 5=2+2+1$ or $5=4+1$
Therefore we can rewrite ${{i}^{5}}$, as
$\Rightarrow {{i}^{5}}={{i}^{(2+2+1)}}$
$\Rightarrow {{i}^{5}}={{i}^{2}}\times {{i}^{2}}\times i$
$\Rightarrow {{i}^{5}}=(-1)\times (-1)\times \sqrt{-1}$
$\Rightarrow {{i}^{5}}=\sqrt{-1}$
$\Rightarrow {{i}^{5}}=i$

Hence the simplified value of ${{i}^{5}}$is $i$.

Note:
If we add two complex numbers then each part will be added separately means the imaginary part will be added with the imaginary and the real part will be added with the real part.
For example: Add the following complex number ${{z}_{1}}=1+6i$and ${{z}_{2}}=-2+0i$
The solution will be ${{z}_{1}}+{{z}_{2}}=1+6i+(-2+0i)$
$\Rightarrow 1+(-2)+(6+0)i$
$\Rightarrow -1+6i$.