Question

What is the value of ${{i}^{34}}$?

Answer 1

For solving this question you should know about the value of ${{i}^{2}}$. If we want to solve this then we can make the factor form of this and we can also write it as a power of ${{i}^{2}}$. And if the power is multiplied with that then the answer will be determined for this.

Complete step-by-step answer:
As our question asked us to determine the value of ${{i}^{34}}$.
As we know that the value of ${{i}^{2}}=-1$ which is negative value and we will write our term ${{i}^{34}}$ as a form of power of ${{i}^{2}}$.
So, for this we will do factors of ${{i}^{34}}$ and then we will solve this.
Now, if we assume that ${{i}^{2}}=t$ then we can write ${{i}^{34}}={{\left( {{i}^{2}} \right)}^{17}}$ and ${{i}^{2}}=t$.
So, it can be written as: ${{i}^{34}}={{t}^{17}}$ and we know that ${{i}^{2}}=-1$.
So, the value of $t=-1$.
We can write it as: ${{i}^{34}}={{\left( -1 \right)}^{17}}$
As we know that the even power of -1 gives and 1 and odd power of -1 gives us the value -1.
So, here the power is 17 which is an odd number.
So, the value of ${{i}^{34}}=-1$.

Note: Alternate method:
We can check this value or we can determine the value of ${{i}^{34}}$ by this method also:
We can write ${{i}^{34}}$ in form of ${{i}^{2}}$ as ${{\left( {{i}^{2}} \right)}^{17}}$ and it is equal to:
$=\left( {{i}^{2}} \right)\times \left( {{i}^{2}} \right)\times \left( {{i}^{2}} \right)\times \left( {{i}^{2}} \right)\times \left( {{i}^{2}} \right)\times \left( {{i}^{2}} \right)\times \left( {{i}^{2}} \right)\times \left( {{i}^{2}} \right)\times \left( {{i}^{2}} \right)\times \left( {{i}^{2}} \right)\times \left( {{i}^{2}} \right)\times \left( {{i}^{2}} \right)\times \left( {{i}^{2}} \right)\times \left( {{i}^{2}} \right)\times \left( {{i}^{2}} \right)\times \left( {{i}^{2}} \right)\times \left( {{i}^{2}} \right)$
And we know that ${{i}^{2}}=-1$
So, it can be written as:
$=\left( -1 \right)\times \left( -1 \right)\times \left( -1 \right)\times \left( -1 \right)\times \left( -1 \right)\times \left( -1 \right)\times \left( -1 \right)\times \left( -1 \right)\times \left( -1 \right)\times \left( -1 \right)\times \left( -1 \right)\times \left( -1 \right)\times \left( -1 \right)\times \left( -1 \right)\times \left( -1 \right)\times \left( -1 \right)\times \left( -1 \right)$
So, the solution of this is:

& =\left( 1 \right)\times \left( 1 \right)\times \left( 1 \right)\times \left( 1 \right)\times \left( 1 \right)\times \left( 1 \right)\times \left( 1 \right)\times \left( 1 \right)\times \left( -1 \right)\times \left( 1 \right)\times \left( -1 \right) \\\ & =-1 \\\ \end{aligned}$$ So, the value of $${{i}^{34}}$$ is equal to -1. During solving this type questions you should be change your given term in a form of $${{\left( {{i}^{2}} \right)}^{n}}$$ and then if the n is equal to any odd number then the value of that is -1 but if the n is equal to even number then the value of that term will be equal to 1.