How do you simplify ${{i}^{28}}$?

Hint: To solve these types of questions we need to know about iota and its properties. Iota is a complex number (A number of the form z = x + iy, where x, y ∈ R, is called a complex number.) which is denoted by $i$ and $i=\sqrt{-1}$.

Complete step by step solution:
For solving this question we should know about Complex numbers and some of the properties of iota.
Iota is a complex number that is denoted by $i$ and the value of iota is $i=\sqrt{-1}$.
Complex number is the number of the form \[z\text{ }=\text{ }x\text{ }+\text{ }iy\], where \[x,\text{ }y\in ~R\]. The numbers $x$ and $y$ are called respectively real and imaginary parts of complex numbers $z$. Which is \[x\text{ }=\text{ }Re\text{ }\left( z \right)\text{ }and\text{ }y\text{ }=\text{ }Im\text{ }\left( z \right)\].
As we know,
$i=\sqrt{-1}$
Therefore we can say that , If we square both sides of the above equation, we get
${{i}^{2}}=-1$
And on again squaring on both sides we get,
 ${{i}^{4}}=1$
Hence now for ${{i}^{28}}$,
${{i}^{28}}={{\left( {{i}^{4}} \right)}^{7}}={{\left( 1 \right)}^{7}}=1$

Hence, On simplifying ${{i}^{28}}$ will be equivalent to $1$.

Note:
On solving such questions the student should pay special attention to breaking up the powers of $i$ so that it can correlate with either multiple of $1,2,3 \text{ and }4$and then we can get the solution.