
How do you simplify ${{i}^{7}}$?
Answer
453.3k+ views
Hint: Since $i$ is defined as the square root of negative of one, so its square root will be equal to one, that is, ${{i}^{2}}=-1$. On squaring this equation we will obtain ${{i}^{4}}=1$. On further squaring, we will obtain ${{i}^{8}}=1$. The given expression is to be multiplied and divided by $i$ to get $\dfrac{{{i}^{8}}}{i}$ which will be simplified to $\dfrac{1}{i}$ after substituting ${{i}^{8}}=1$. Then finally multiplying and dividing the numerator by $i$ and substituting ${{i}^{2}}=-1$ we will get the final simplified expression.
Complete step by step solution:
We know that $i$ is defined as
\[\Rightarrow i=\sqrt{-1}\]
Taking the square of both the sides, we get
$\Rightarrow {{i}^{2}}=-1........\left( i \right)$
Again taking square on both the sides of the above equation we get
$\Rightarrow {{i}^{4}}=1........\left( ii \right)$
Now, let us write the given expression as
$\Rightarrow E={{i}^{7}}$
Multiplying and dividing by $i$, we get
$\begin{align}
& \Rightarrow E=\dfrac{{{i}^{7}}\times i}{i} \\
& \Rightarrow E=\dfrac{{{i}^{8}}}{i} \\
\end{align}$
The numerator of the above expression can also be written as
$\Rightarrow E=\dfrac{{{\left( {{i}^{4}} \right)}^{2}}}{i}$
Substituting (ii) in the above expression, we get
$\begin{align}
& \Rightarrow E=\dfrac{{{\left( 1 \right)}^{2}}}{i} \\
& \Rightarrow E=\dfrac{1}{i} \\
\end{align}$
Again multiplying and dividing by $i$, we get
$\Rightarrow E=\dfrac{i}{{{i}^{2}}}$
Finally, substituting (i) in the above expression, we get
$\begin{align}
& \Rightarrow E=\dfrac{i}{-1} \\
& \Rightarrow E=-i \\
\end{align}$
Hence, the given expression is simplified as $-i$.
Note: Instead of multiplying and dividing the given expression by $i$, we can also extract the highest power of ${{i}^{2}}$ from the given expression. In doing so, we will obtain the given expression as ${{\left( {{i}^{2}} \right)}^{3}}i$. Then, as we know that ${{i}^{2}}=-1$, the expression will get reduced to ${{\left( -1 \right)}^{3}}i$ which will be simplified to $-i$ which we have obtained in the above solution.
Complete step by step solution:
We know that $i$ is defined as
\[\Rightarrow i=\sqrt{-1}\]
Taking the square of both the sides, we get
$\Rightarrow {{i}^{2}}=-1........\left( i \right)$
Again taking square on both the sides of the above equation we get
$\Rightarrow {{i}^{4}}=1........\left( ii \right)$
Now, let us write the given expression as
$\Rightarrow E={{i}^{7}}$
Multiplying and dividing by $i$, we get
$\begin{align}
& \Rightarrow E=\dfrac{{{i}^{7}}\times i}{i} \\
& \Rightarrow E=\dfrac{{{i}^{8}}}{i} \\
\end{align}$
The numerator of the above expression can also be written as
$\Rightarrow E=\dfrac{{{\left( {{i}^{4}} \right)}^{2}}}{i}$
Substituting (ii) in the above expression, we get
$\begin{align}
& \Rightarrow E=\dfrac{{{\left( 1 \right)}^{2}}}{i} \\
& \Rightarrow E=\dfrac{1}{i} \\
\end{align}$
Again multiplying and dividing by $i$, we get
$\Rightarrow E=\dfrac{i}{{{i}^{2}}}$
Finally, substituting (i) in the above expression, we get
$\begin{align}
& \Rightarrow E=\dfrac{i}{-1} \\
& \Rightarrow E=-i \\
\end{align}$
Hence, the given expression is simplified as $-i$.
Note: Instead of multiplying and dividing the given expression by $i$, we can also extract the highest power of ${{i}^{2}}$ from the given expression. In doing so, we will obtain the given expression as ${{\left( {{i}^{2}} \right)}^{3}}i$. Then, as we know that ${{i}^{2}}=-1$, the expression will get reduced to ${{\left( -1 \right)}^{3}}i$ which will be simplified to $-i$ which we have obtained in the above solution.
Recently Updated Pages
The correct geometry and hybridization for XeF4 are class 11 chemistry CBSE

Water softening by Clarks process uses ACalcium bicarbonate class 11 chemistry CBSE

With reference to graphite and diamond which of the class 11 chemistry CBSE

A certain household has consumed 250 units of energy class 11 physics CBSE

The lightest metal known is A beryllium B lithium C class 11 chemistry CBSE

What is the formula mass of the iodine molecule class 11 chemistry CBSE

Trending doubts
Is Cellular respiration an Oxidation or Reduction class 11 chemistry CBSE

In electron dot structure the valence shell electrons class 11 chemistry CBSE

What is the Pitti Island famous for ABird Sanctuary class 11 social science CBSE

State the laws of reflection of light

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells
