Question

How do you simplify ${{i}^{37}}$?

Answer 1

In this question, we have to find ${{i}^{37}}$. In solving this question, we will first know about the term$i$. After that, we will get to know what will happen to the if we put powers to the $i$. After that, we will solve the question.

Complete answer:
Let us solve this question.
In this question, we have just asked to simplify the value of ${{i}^{37}}$.
Let us first know about $i$.
$i$ is read as iota. $i$ is a complex number. It is an imaginary number whose value is $\sqrt{-1}$. That means iota is the square root of one negative of one.
Now, let us find the value of the square of iota.
${{i}^{2}}=i\times i=\sqrt{-1}\times \sqrt{-1}$
As we know that $\sqrt{a}\times \sqrt{a}=a$. So, we can write the above equation as
$\Rightarrow {{i}^{2}}=-1$
So, we get that the square of iota is negative of 1.
Now, let us find the cube of iota.
${{i}^{3}}={{i}^{2}}\times i$
As we know that square of iota is -1. So, we can write the above equation as
$\Rightarrow {{i}^{3}}=-1\times i=-i$
Hence, we get that the cube of iota is negative of iota.
Now, let us find the iota having power 4.
As we can write ${{i}^{4}}={{i}^{2}}\times {{i}^{2}}$.
And we know that the square of iota is -1. Using this in above equation, we get
$\Rightarrow {{i}^{4}}=(-1)\times (-1)=1$
Now, we can say that

& {{i}^{1}}=i \\\ & {{i}^{2}}=-1 \\\ & {{i}^{3}}=-i \\\ & {{i}^{4}}=1 \\\ \end{aligned}$$ Similarly, we can write $${{i}^{5}}={{i}^{4}}\times i=1\times i=i$$ $${{i}^{6}}={{i}^{4}}\times {{i}^{2}}=1\times (-1)=-1$$ $${{i}^{7}}={{i}^{4}}\times {{i}^{3}}=1\times (-i)=-i$$ $${{i}^{8}}={{i}^{4}}\times {{i}^{4}}=1\times 1=1$$ So, we can say that from the above equations $${{i}^{4n+1}}=i$$ $${{i}^{4n+2}}=-1$$ $${{i}^{4n+3}}=-i$$ $${{i}^{4n+4}}=1$$ By using the above equations, we can find the value of $${{i}^{37}}$$. As we can write 37 as 4n+1. So, $${{i}^{37}}$$ will be equal to $$i$$. **Note:** For solving this question, we should have a better knowledge in pre-calculus. Always remember the formulas which are given in the following. $$\begin{aligned} & {{i}^{1}}=i \\\ & {{i}^{2}}=-1 \\\ & {{i}^{3}}=-i \\\ & {{i}^{4}}=1 \\\ \end{aligned}$$ And $${{i}^{4n+1}}=i$$ $${{i}^{4n+2}}=-1$$ $${{i}^{4n+3}}=-i$$ $${{i}^{4n+4}}=1$$ To solve this type of question easily, remember the above formulas.