Question

If n is any positive integer, find the value of $\dfrac{{{i}^{4n+1}}-{{i}^{4n-1}}}{2}$.

Answer 1

Hint: Use the fact that $i=\sqrt{-1}$ is a square root of unity. Calculate higher powers of $i$ and simplify the given expression using the law of exponents. Then substitute the values of higher powers of $i$ to calculate the value of the given expression.

Complete step-by-step solution -
We have to calculate the value of $\dfrac{{{i}^{4n+1}}-{{i}^{4n-1}}}{2}$. We observe that an expression is a complex number.
We know that $i=\sqrt{-1}$. We will now calculate higher powers of $i$.
Thus, we have ${{i}^{2}}={{\left( \sqrt{-1} \right)}^{2}}=-1,{{i}^{3}}={{i}^{2}}\times i=-i,{{i}^{4}}={{\left( {{i}^{2}} \right)}^{2}}=1$.
We know the laws of exponents state that ${{a}^{b}}\times {{a}^{c}}={{a}^{b+c}}$ and ${{\left( {{a}^{b}} \right)}^{c}}={{a}^{bc}}$.
So, we can simplify the expression ${{i}^{4n+1}}$ as ${{i}^{4n+1}}={{i}^{4n}}\times i={{\left( {{i}^{4}} \right)}^{n}}\times i$.
We know that ${{i}^{4}}=1$.
Thus, we have ${{i}^{4n+1}}={{i}^{4n}}\times i={{\left( {{i}^{4}} \right)}^{n}}\times i={{1}^{n}}\times i=1\times i=i.....\left( 1 \right)$.
We will now simplify the expression ${{i}^{4n-1}}$.
Using the laws of exponents, we can rewrite the expression ${{i}^{4n-1}}$ as ${{i}^{4n-1}}={{i}^{4n}}\times {{i}^{-1}}={{\left( {{i}^{4}} \right)}^{n}}\times {{i}^{-1}}$.
We know that ${{i}^{4}}=1$.
Thus, we have ${{i}^{4n-1}}={{i}^{4n}}\times {{i}^{-1}}={{\left( {{i}^{4}} \right)}^{n}}\times {{i}^{-1}}={{1}^{n}}\times {{i}^{-1}}=1\times \dfrac{1}{i}=\dfrac{1}{i}$.
We will further simplify this expression by multiplying and dividing it by $i$.
Thus, we have ${{i}^{4n-1}}=\dfrac{1}{i}=\dfrac{1}{i}\times \dfrac{i}{i}$.
Simplifying the above expression, we have ${{i}^{4n-1}}=\dfrac{1}{i}=\dfrac{1}{i}\times \dfrac{i}{i}=\dfrac{i}{{{i}^{2}}}$.
We know that ${{i}^{2}}=-1$.
Thus, we have ${{i}^{4n-1}}=\dfrac{1}{i}=\dfrac{1}{i}\times \dfrac{i}{i}=\dfrac{i}{{{i}^{2}}}=\dfrac{i}{-1}=-i.....\left( 2 \right)$.
Substituting equation (1) and (2) in the expression $\dfrac{{{i}^{4n+1}}-{{i}^{4n-1}}}{2}$, we have $\dfrac{{{i}^{4n+1}}-{{i}^{4n-1}}}{2}=\dfrac{i-\left( -i \right)}{2}$.
Thus, we have $\dfrac{{{i}^{4n+1}}-{{i}^{4n-1}}}{2}=\dfrac{i-\left( -i \right)}{2}=\dfrac{i+i}{2}=\dfrac{2i}{2}=i$.
Hence, the value of the expression $\dfrac{{{i}^{4n+1}}-{{i}^{4n-1}}}{2}$ is $i$ for all positive integer values of n.

Note: We can also solve this question by applying induction on ‘n’. We can substitute any positive integer value of n; the value of the expression won’t change. We can write any complex number in the form $a+ib$, where $ib$ is the imaginary part and $a$ is the real part.