Question

Revaluate the following: \[{{\left[ {{i}^{18}}+{{\left( \dfrac{1}{i} \right)}^{25}} \right]}^{3}}\].

Answer 1

Hint: In the above question we will use the properties of iota (i) given as follows:
\[\begin{align}
  & {{i}^{4n}}=1 \\
 & {{i}^{4n+1}}=i \\
 & {{i}^{4n+2}}=-1 \\
 & {{i}^{4n+3}}=-i \\
\end{align}\]
where n can be 0, 1, 2, 3…

Complete step-by-step answer:
We have been given the expression \[{{\left[ {{i}^{18}}+{{\left( \dfrac{1}{i} \right)}^{25}} \right]}^{3}}\].
Now we will rationalize the term \[\left( \dfrac{1}{i} \right)\], i.e. we will multiply the numerator as well as the denominator by iota (i).
\[\Rightarrow \dfrac{1}{i}\times \dfrac{i}{i}=\dfrac{i}{{{i}^{2}}}\]
And we know that \[{{i}^{2}}=-1\].
\[\Rightarrow \dfrac{1}{i}\times \dfrac{i}{-1}=-1\]
So by substituting the value of \[\left( \dfrac{1}{i} \right)\] in the above expression, we get as follows:
\[{{\left[ {{i}^{18}}+{{\left( \dfrac{1}{i} \right)}^{25}} \right]}^{3}}={{\left[ {{i}^{18}}+{{\left( -i \right)}^{25}} \right]}^{3}}\]
We know that if the power of iota (i) is in the form of \[{{i}^{4n+2}}\], then the value of \[{{i}^{4n+2}}=-1\].
\[\Rightarrow {{i}^{18}}={{i}^{\left( 4\times 4+2 \right)}}=-1\]
We also know that if the power of iota (i) is in the form of \[{{i}^{4n+1}}\], then the value of \[{{i}^{4n+1}}=i\].
\[\Rightarrow {{i}^{25}}={{i}^{\left( 4\times 6+1 \right)}}=i\]
So on substituting the values of \[{{i}^{18}}\] and \[{{i}^{25}}\] in the above equation, we get as follows:
\[{{\left[ {{i}^{18}}+{{\left( \dfrac{1}{i} \right)}^{25}} \right]}^{3}}={{\left[ -1+{{\left( -1 \right)}^{25}}i \right]}^{3}}\]
Also, we know that \[{{\left( -1 \right)}^{25}}=-1\] so by substituting the value in the above equation, we get as follows:
\[{{\left[ {{i}^{18}}+{{\left( \dfrac{1}{i} \right)}^{25}} \right]}^{3}}={{\left[ -1+i \right]}^{3}}={{\left( -1 \right)}^{3}}{{\left[ 1-i \right]}^{3}}\]
We already know the identity \[{{\left( a+b \right)}^{3}}={{a}^{3}}+{{b}^{3}}+3ab\left( a+b \right)\].
So by using this identity in the above expression, we get as follows:
\[{{\left[ {{i}^{18}}+{{\left( \dfrac{1}{i} \right)}^{25}} \right]}^{3}}={{\left( -1 \right)}^{3}}{{\left[ 1+i \right]}^{3}}={{\left( -1 \right)}^{3}}\left[ {{1}^{3}}+{{i}^{3}}+3i\left( 1+i \right) \right]={{\left( -1 \right)}^{3}}\left[ 1+{{i}^{3}}+3i+3{{i}^{2}} \right]\]
We know that \[{{\left( -1 \right)}^{3}}=-1\].
So, using this in the above equation, we get as follows:
\[{{\left[ {{i}^{18}}+{{\left( \dfrac{1}{i} \right)}^{25}} \right]}^{3}}=\left( -1 \right)\left[ 1+{{i}^{3}}+3i+3{{i}^{2}} \right]\]
We also know that \[{{i}^{3}}=-i\] and \[{{i}^{2}}=-1\].
So by using these values in the above equation, we get as follows:
\[{{\left[ {{i}^{18}}+{{\left( \dfrac{1}{i} \right)}^{25}} \right]}^{3}}=-\left[ 1-i+3i+3(-1) \right]=-\left[ 1+2i-3 \right]=-\left[ 2i-2 \right]=2\left( 1-i \right)\]
Hence the value of the given expression \[{{\left[ {{i}^{18}}+{{\left( \dfrac{1}{i} \right)}^{25}} \right]}^{3}}\] is \[2\left( 1-i \right)\].

Note: Be careful while finding the value of \[{{i}^{18}}\] and \[{{i}^{25}}\]. Also take care of the sign while expanding \[{{\left( 1+i \right)}^{3}}\] using the identity. Also, remember that ‘i’ is known as iota and is equal to \[\sqrt{-1}\] which is an imaginary number.