Question

How do you simplify \[{{i}^{37}}\]?

Answer 1

Hint: 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
\[\begin{align}
  & {{i}^{1}}=i \\
 & {{i}^{2}}=-1 \\
 & {{i}^{3}}=-i \\
 & {{i}^{4}}=1 \\
\end{align}\]
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.
\[\begin{align}
  & {{i}^{1}}=i \\
 & {{i}^{2}}=-1 \\
 & {{i}^{3}}=-i \\
 & {{i}^{4}}=1 \\
\end{align}\]
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.