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Question

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

How do you simplify i37{{i}^{37}}?

Explanation

Solution

In this question, we have to find i37{{i}^{37}}. In solving this question, we will first know about the termii. After that, we will get to know what will happen to the if we put powers to the ii. 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 i37{{i}^{37}}.
Let us first know about ii.
ii is read as iota. ii is a complex number. It is an imaginary number whose value is 1\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.
i2=i×i=1×1{{i}^{2}}=i\times i=\sqrt{-1}\times \sqrt{-1}
As we know that a×a=a\sqrt{a}\times \sqrt{a}=a. So, we can write the above equation as
i2=1\Rightarrow {{i}^{2}}=-1
So, we get that the square of iota is negative of 1.
Now, let us find the cube of iota.
i3=i2×i{{i}^{3}}={{i}^{2}}\times i
As we know that square of iota is -1. So, we can write the above equation as
i3=1×i=i\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 i4=i2×i2{{i}^{4}}={{i}^{2}}\times {{i}^{2}}.
And we know that the square of iota is -1. Using this in above equation, we get
i4=(1)×(1)=1\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.