Question

How do you simplify ${{i}^{27}}$?

Answer 1

Hint: The number $i$ is the symbol for the square root of negative of unity, a complex number. Its square is equal to the negative of unity, that is, ${{i}^{2}}=-1$. On multiplying with $i$, we get ${{i}^{3}}=-i$. We can write the exponent of $27$ raised on $i$ in the given expression as $3\times 9$ so that the given expression will become ${{\left( {{i}^{3}} \right)}^{^{9}}}$ using the exponent property \[{{a}^{mn}}={{\left( {{a}^{m}} \right)}^{n}}\]. Then using the derived relation ${{i}^{3}}=-i$, the expression will be reduced to \[{{\left( -i \right)}^{9}}\]. On further solving the expression by using the property ${{\left( ab \right)}^{m}}={{a}^{m}}{{b}^{m}}$ and the relation ${{i}^{3}}=-i$, we will get the final simplified expression.

Complete step by step solution:
We know that $i$ is a complex number, which is equal to the square root of negative of unity, that is,
$\Rightarrow i=\sqrt{-1}........\left( i \right)$
Squaring both the sides, we get
$\begin{align}
  & \Rightarrow {{i}^{2}}={{\left( \sqrt{-1} \right)}^{2}} \\
 & \Rightarrow {{i}^{2}}=-1........\left( ii \right) \\
\end{align}$
Multiplying the equations (i) and (ii) we get
\[\begin{align}
  & \Rightarrow i\times {{i}^{2}}=\sqrt{-1}\times -1 \\
 & \Rightarrow {{i}^{3}}=i\times -1 \\
 & \Rightarrow {{i}^{3}}=-i.........\left( iii \right) \\
\end{align}\]
Now, we consider the expression given in the above question is
$\Rightarrow E={{i}^{27}}$
Now, we know that $27=3\times 9$. So the above expression can be written as
$\Rightarrow E={{i}^{3\times 9}}$
Now, we know the exponent property \[{{a}^{mn}}={{\left( {{a}^{m}} \right)}^{n}}\]. Therefore, the above expression can also be written as
$\Rightarrow E={{\left( {{i}^{3}} \right)}^{9}}$
Substituting (iii) we get
$\Rightarrow E={{\left( -i \right)}^{9}}$
Using the exponent property ${{\left( ab \right)}^{m}}={{a}^{m}}{{b}^{m}}$ we can write
$\begin{align}
  & \Rightarrow E={{\left( -1 \right)}^{9}}{{i}^{9}} \\
 & \Rightarrow E=-{{i}^{9}} \\
\end{align}$
Now, writing $9=3\times 3$ in the above expression we get
\[\Rightarrow E=-{{i}^{3\times 3}}\]
Again using the property \[{{a}^{mn}}={{\left( {{a}^{m}} \right)}^{n}}\] we get
\[\Rightarrow E=-{{\left( {{i}^{3}} \right)}^{3}}\]
Again substituting (iii) we get
\[\begin{align}
  & \Rightarrow E=-{{\left( -i \right)}^{3}} \\
 & \Rightarrow E=-{{\left( -1 \right)}^{3}}{{i}^{3}} \\
 & \Rightarrow E=-\left( -1 \right){{i}^{3}} \\
 & \Rightarrow E={{i}^{3}} \\
\end{align}\]
Finally, substituting (iii) we get
$\Rightarrow E=-i$

Hence, the given expression is simplified as $-i$.

Note: We can also attempt this question very easily by using the relation ${{i}^{4}}=1$. For using this relation, we need to multiply and divide the given expression by $i$ to get \[\dfrac{{{i}^{28}}}{i}\] which can be written as $\dfrac{{{\left( {{i}^{4}} \right)}^{7}}}{i}$. Substituting ${{i}^{4}}=1$ in this, we will obtain the given expression as $\dfrac{1}{i}$ which will be equal to $-i$.