
How do you simplify ${{i}^{13}}$ ?
Answer
530.1k+ views
Hint: From the question given we have to simplify or we have to find the value of ${{i}^{13}}$. As we know that the values of $i=i$, ${{i}^{2}}=-1$, ${{i}^{3}}=-i$, ${{i}^{4}}=1$ and further multiplying with i repeats the same pattern. From this we can generalize that the let ${{i}^{n}}$ the power n is divided with 4 if it leaves remainder 1 then the value of ${{i}^{n}}$ is i, if it leaves remainder 2 then the value of ${{i}^{n}}$ is -1, if it leaves the remainder 3 then the ${{i}^{n}}$ is -i, if the remainder is zero then the value of ${{i}^{n}}$ is 1. From this we will get the value of ${{i}^{13}}$.
Complete step by step solution:
From the question given we have to simplify the
$\Rightarrow {{i}^{13}}$
As we know that the values of
$\begin{align}
& \Rightarrow i=i \\
& \Rightarrow {{i}^{2}}=-1 \\
& \Rightarrow {{i}^{3}}=-i \\
& \Rightarrow {{i}^{4}}=1 \\
\end{align}$
And further multiplying with i we will get the same repeated pattern.
From this we can conclude that
let ${{i}^{n}}$ now the power n is divided with 4
if the remainder is equals to 1 then the value of
$\Rightarrow {{i}^{n}}=i$
if the remainder is equals to 2 then the value of
$\Rightarrow {{i}^{n}}=-1$
if the remainder is equals to 3 then the value of
$\Rightarrow {{i}^{n}}=-i$
if the remainder is equals to 0 then the value of
$\Rightarrow {{i}^{n}}=1$
Now here in the question we have to find ${{i}^{13}}$
Divide the 13 with four we will get the remainder as 1
As we know that if the remainder is 1 then the value of ${{i}^{13}}$is
$\Rightarrow {{i}^{13}}={{i}^{1}}=i$
Therefore, the simplification is $\Rightarrow {{i}^{13}}={{i}^{1}}=i$.
Note: students should generalize to solve these types of question it will make easy to solve, students should also know that the value of square root of -1 is $\sqrt{-1}=\sqrt{{{i}^{2}}}=i$.
We can also solve this question as follows.
\[\Rightarrow {{i}^{13}}={{i}^{12}}\times i\]
\[\Rightarrow {{i}^{13}}={{i}^{4}}\times {{i}^{4}}\times {{i}^{4}}\times i\]
We know that \[{{i}^{4}}=1\],
\[\Rightarrow {{i}^{13}}=1\times 1\times 1\times i\]
Therefore,
\[\Rightarrow {{i}^{13}}=i\].
Complete step by step solution:
From the question given we have to simplify the
$\Rightarrow {{i}^{13}}$
As we know that the values of
$\begin{align}
& \Rightarrow i=i \\
& \Rightarrow {{i}^{2}}=-1 \\
& \Rightarrow {{i}^{3}}=-i \\
& \Rightarrow {{i}^{4}}=1 \\
\end{align}$
And further multiplying with i we will get the same repeated pattern.
From this we can conclude that
let ${{i}^{n}}$ now the power n is divided with 4
if the remainder is equals to 1 then the value of
$\Rightarrow {{i}^{n}}=i$
if the remainder is equals to 2 then the value of
$\Rightarrow {{i}^{n}}=-1$
if the remainder is equals to 3 then the value of
$\Rightarrow {{i}^{n}}=-i$
if the remainder is equals to 0 then the value of
$\Rightarrow {{i}^{n}}=1$
Now here in the question we have to find ${{i}^{13}}$
Divide the 13 with four we will get the remainder as 1
As we know that if the remainder is 1 then the value of ${{i}^{13}}$is
$\Rightarrow {{i}^{13}}={{i}^{1}}=i$
Therefore, the simplification is $\Rightarrow {{i}^{13}}={{i}^{1}}=i$.
Note: students should generalize to solve these types of question it will make easy to solve, students should also know that the value of square root of -1 is $\sqrt{-1}=\sqrt{{{i}^{2}}}=i$.
We can also solve this question as follows.
\[\Rightarrow {{i}^{13}}={{i}^{12}}\times i\]
\[\Rightarrow {{i}^{13}}={{i}^{4}}\times {{i}^{4}}\times {{i}^{4}}\times i\]
We know that \[{{i}^{4}}=1\],
\[\Rightarrow {{i}^{13}}=1\times 1\times 1\times i\]
Therefore,
\[\Rightarrow {{i}^{13}}=i\].
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