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:

& {{i}^{4n}}=1 \\\ & {{i}^{4n+1}}=i \\\ & {{i}^{4n+2}}=-1 \\\ & {{i}^{4n+3}}=-i \\\ \end{aligned}$$ where n can be 0, 1, 2, 3… _Complete step-by-step answer:_ We have been given the expression $${{\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.