Question

How do you simplify ${i^{14}}?$

Answer 1

We will use the formula of imaginary number to simplify the given question.
On doing some simplification we get the required answer.

Formula Used:
In mathematics, if any negative number is written under the square root then it is called an imaginary number.
The imaginary unit number is called as complex numbers, where $i$ is defined as imaginary or unit imaginary.
Suppose we want to calculate roots for the equation ${x^2} = 1$ , there we can find two different real roots of $x$ .
But if we want to find roots of $x$ in the equation ${x^3} = 1$ , there we can have three different roots for $x$ .
So, to solve it, we can perform following steps:
${x^3} = 1$
$\Rightarrow {x^3} - 1 = 0$
$\Rightarrow (x - 1)({x^2} + x + 1) = 0$
So, it is true for the above equation that:
Either $(x - 1) = 0$ or $({x^2} + x + 1) = 0$ .
So, $x$ can have a value of $1$ .
And, if we solve $({x^2} + x + 1) = 0$ , we can find two roots of $x$ also.
So,
$\Rightarrow x = \dfrac{{ - 1 \pm \sqrt {{1^2} - 4 \times 1 \times 1} }}{{2 \times 1}} = \dfrac{{ - 1 \pm \sqrt { - 3} }}{2} = \dfrac{{ - 1 \pm \sqrt 3 i}}{2}.$
So, the number $- 1$ , under the square root is called an imaginary unit and this kind of roots or numbers are called complex numbers.
So, the quantity of $i$ is $\sqrt { - 1}$ .
Or we can write it as $i = \sqrt { - 1}$ .
So, it is obvious that if we multiply $i$ even numbers of times then it will give us a real number, but if we multiply $i$ odd numbers of times then it will give us an imaginary number.

Complete step by step answer:
The given expression in the question is ${i^{14}}$ .
We can re-write the power of $i$ in following way:
${i^{14}} = {\left( {{i^2}} \right)^7}...............(1)$
Now, we know that $i = \sqrt { - 1}$ .
So, if we squared both the terms in $i = \sqrt { - 1}$ , we get:
${i^2} = - 1$ .
So, we can re-write the equation $(1)$ as following:
${i^{14}} = {\left( { - 1} \right)^7}$ .
Now, if we multiply $- 1$ even number of times it will give us $1$ but if we multiply $- 1$ odd numbers of times it will give us $- 1$ .
Here, the power of $- 1$ is $7$ , which is an odd number.
Therefore,
${i^{14}} = {\left( { - 1} \right)^7} = - 1$ .

The value of ${i^{14}} = - 1$ .

Note: Points to remember:
A complex number is expressed as following:
$X + i.Y$ , where $X$ and $Y$ are real numbers but the imaginary part of the number is $i$ .
A complex number lies on the imaginary axis in $X - Y$ plane.