Question

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

Answer 1

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
$\begin{aligned} & \Rightarrow {{i}^{2}}={{\left( \sqrt{-1} \right)}^{2}} \\\ & \Rightarrow {{i}^{2}}=-1........\left( ii \right) \\\ \end{aligned}$
Multiplying the equations (i) and (ii) we get

& \Rightarrow i\times {{i}^{2}}=\sqrt{-1}\times -1 \\\ & \Rightarrow {{i}^{3}}=i\times -1 \\\ & \Rightarrow {{i}^{3}}=-i.........\left( iii \right) \\\ \end{aligned}$$ Now, we consider the expression given in the above question is $\Rightarrow E={{i}^{27}}$ Now, we know that $27=3\times 9$. So the above expression can be written as $\Rightarrow E={{i}^{3\times 9}}$ Now, we know the exponent property $${{a}^{mn}}={{\left( {{a}^{m}} \right)}^{n}}$$. Therefore, the above expression can also be written as $\Rightarrow E={{\left( {{i}^{3}} \right)}^{9}}$ Substituting (iii) we get $\Rightarrow E={{\left( -i \right)}^{9}}$ Using the exponent property ${{\left( ab \right)}^{m}}={{a}^{m}}{{b}^{m}}$ we can write $\begin{aligned} & \Rightarrow E={{\left( -1 \right)}^{9}}{{i}^{9}} \\\ & \Rightarrow E=-{{i}^{9}} \\\ \end{aligned}$ Now, writing $9=3\times 3$ in the above expression we get $$\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{aligned} & \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{aligned}$$ Finally, substituting (iii) we get $\Rightarrow E=-i$ **Hence, the given expression is simplified as $-i$.** **Note:** We can also attempt this question very easily by using the relation ${{i}^{4}}=1$. For using this relation, we need to multiply and divide the given expression by $i$ to get $$\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 $-i$.