Question

Find the value of $\dfrac{{{i^{4n + 1}} - {i^{4n - 1}}}}{2}$.
A) $–1$
B) $1$
C) $–i$
D) $i$

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 $i$ and use laws of exponents 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
${i^2} = {\left( {\sqrt { - 1} } \right)^2}$
Square the term,
$ \Rightarrow {i^2} = - 1$
Now find the value of ${i^3}$,
$ \Rightarrow {i^3} = {i^2} \times i$
Substitute the value of ${i^2}$,
$ \Rightarrow {i^3} = - i$
Now find the value of ${i^4}$,
$ \Rightarrow {i^4} = {\left( {{i^2}} \right)^2}$
Substitute the value of ${i^2}$,
$ \Rightarrow {i^4} = {\left( { - 1} \right)^2}$
Square the term,
$ \Rightarrow {i^4} = 1$
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,
$ \Rightarrow {i^{4n + 1}} = {i^{4n}} \times i$
Use the exponent law,
$ \Rightarrow {i^{4n + 1}} = {\left( {{i^4}} \right)^n} \times i$
Substitute the value of ${i^4}$,
$ \Rightarrow {i^{4n + 1}} = {1^n} \times i$
As we know that any power of 1 returns 1.
$ \Rightarrow {i^{4n + 1}} = i$.................….. (1)
Now, we can simplify the expression ${i^{4n - 1}}$ as,
$ \Rightarrow {i^{4n + 1}} = {i^{4n}} \times {i^{ - 1}}$
Use the exponent law,
$ \Rightarrow {i^{4n - 1}} = \dfrac{{{{\left( {{i^4}} \right)}^n}}}{i}$
Substitute the value of ${i^4}$,
$ \Rightarrow {i^{4n - 1}} = \dfrac{{{1^n}}}{i}$
As we know that any power of 1 returns 1.
$ \Rightarrow {i^{4n - 1}} = \dfrac{1}{i}$
Rationalize the term by multiplying numerator and denominator by $i$,
$ \Rightarrow {i^{4n - 1}} = \dfrac{1}{i} \times \dfrac{i}{i}$
Multiply the terms,
$ \Rightarrow {i^{4n - 1}} = \dfrac{i}{{{i^2}}}$
Substitute the value of ${i^2}$,
$ \Rightarrow {i^{4n - 1}} = \dfrac{i}{{ - 1}}$
Simplify the terms,
$ \Rightarrow {i^{4n - 1}} = - i$..............….. (2)
Substitute the values from equation (1) and (2) in original expression,
$ \Rightarrow \dfrac{{{i^{4n + 1}} - {i^{4n - 1}}}}{2} = \dfrac{{i - \left( { - i} \right)}}{2}$
Open the bracket and change the sign accordingly,
$ \Rightarrow \dfrac{{{i^{4n + 1}} - {i^{4n - 1}}}}{2} = \dfrac{{i + i}}{2}$
Add the term in the numerator,
$ \Rightarrow \dfrac{{{i^{4n + 1}} - {i^{4n - 1}}}}{2} = \dfrac{{2i}}{2}$
Cancel out the common factor,
$\therefore \dfrac{{{i^{4n + 1}} - {i^{4n - 1}}}}{2} = i$
So, the value of $\dfrac{{{i^{4n + 1}} - {i^{4n - 1}}}}{2}$ is $i$.

Hence, the option (D) is the correct answer.

Note: The complex numbers are the field C of numbers of the form $x + iy$, where x and y are real numbers and i is the imaginary unit equal to the square root of -1. When a single letter z is used to denote a complex number. It is denoted as $z = x + iy$.