Question

What does ${i^4}$ equal?

Answer 1

Hint: In this question we know the value of $i\left( {iota} \right)$ as $\sqrt { - 1} $. First, we need to find the value of its square and then we will find the will of iota to raise to the power four. We know that iota is a complex number or we can say that it is an imaginary number.

Complete step by step answer:
In the above question, we know that \[\;i = \sqrt { - 1} .\]
Therefore, we need to find \[{i^4} = {\left( {\sqrt { - 1} } \right)^4}\]
When we have a number, say $\sqrt 2 $ and we multiply it by another $\sqrt 2 $, we get what's inside the square root sign:
$\sqrt 2 \times \sqrt 2 = 2$
So, let's apply that to our problem:
\[{\left( {\sqrt { - 1} } \right)^2} = - 1\]
But our problem has four $i$, not two. So, let's break it down into two sets of two:
$ \Rightarrow {\sqrt { - 1} ^4} = {\sqrt { - 1} ^{2 + 2}} = {\sqrt { - 1} ^2} \times {\sqrt { - 1} ^2} = - 1 \times - 1 = 1$
We could also have done it this way:
$ \Rightarrow {\sqrt { - 1} ^4} = {\sqrt { - 1} ^{2 \times 2}} = {\left( {{{\sqrt { - 1} }^2}} \right)^2} = {\left( { - 1} \right)^2} = 1$
Therefore, the value of ${i^4}$ is equal to $1$.

Additional information: The imaginary part of a complex number is defined as ‘iota’. To calculate the value of an imaginary number, we use the notation iota or $i$. The square root of a negative number gives us an imaginary number. Value of $i$ is $\sqrt { - 1} $. A negative value inside a square root signifies an imaginary value. All the basic arithmetic operators are applicable to imaginary numbers. On squaring an imaginary number, we obtain a negative value.

Note: The value of higher degrees of i follows a circular formula. Once we reach the power of four of iota, then further values repeat in the same manner. We just only need to know four values as far as the power of iota is concerned.