Solveeit Logo
Login

Question

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

Find the value of i4n+1−i4n−12\dfrac{{{i^{4n + 1}} - {i^{4n - 1}}}}{2}.
A) –1–1
B) 11
C) –i–i
D) ii

Explanation

Solution

Use the fact that i=−1i = \sqrt { - 1} is a square root of unity. Calculate higher powers of ii and simplify the given expression using the law of exponents. Then substitute the values of higher powers of ii 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 i4n+1−i4n−12\dfrac{{{i^{4n + 1}} - {i^{4n - 1}}}}{2}. We observe that an expression is a complex number.
We know that i=−1i = \sqrt { - 1} . We will now calculate higher powers of ii.
Thus, we have
i2=(−1)2{i^2} = {\left( {\sqrt { - 1} } \right)^2}
Square the term,
⇒i2=−1\Rightarrow {i^2} = - 1
Now find the value of i3{i^3},
⇒i3=i2×i\Rightarrow {i^3} = {i^2} \times i
Substitute the value of i2{i^2},
⇒i3=−i\Rightarrow {i^3} = - i
Now find the value of i4{i^4},
⇒i4=(i2)2\Rightarrow {i^4} = {\left( {{i^2}} \right)^2}
Substitute the value of i2{i^2},
⇒i4=(−1)2\Rightarrow {i^4} = {\left( { - 1} \right)^2}
Square the term,
⇒i4=1\Rightarrow {i^4} = 1
The laws of exponents state that,
ab×ac=ab+c{a^b} \times {a^c} = {a^{b + c}}
And,
(ab)c=abc{\left( {{a^b}} \right)^c} = {a^{bc}}
So, we can simplify the expression i4n+1{i^{4n + 1}} as,
⇒i4n+1=i4n×i\Rightarrow {i^{4n + 1}} = {i^{4n}} \times i
Use the exponent law,
⇒i4n+1=(i4)n×i\Rightarrow {i^{4n + 1}} = {\left( {{i^4}} \right)^n} \times i
Substitute the value of i4{i^4},
⇒i4n+1=1n×i\Rightarrow {i^{4n + 1}} = {1^n} \times i
As we know that any power of 1 returns 1.
⇒i4n+1=i\Rightarrow {i^{4n + 1}} = i.................….. (1)
Now, we can simplify the expression i4n−1{i^{4n - 1}} as,
⇒i4n+1=i4n×i−1\Rightarrow {i^{4n + 1}} = {i^{4n}} \times {i^{ - 1}}
Use the exponent law,
⇒i4n−1=(i4)ni\Rightarrow {i^{4n - 1}} = \dfrac{{{{\left( {{i^4}} \right)}^n}}}{i}
Substitute the value of i4{i^4},
⇒i4n−1=1ni\Rightarrow {i^{4n - 1}} = \dfrac{{{1^n}}}{i}
As we know that any power of 1 returns 1.
⇒i4n−1=1i\Rightarrow {i^{4n - 1}} = \dfrac{1}{i}
Rationalize the term by multiplying numerator and denominator by ii,
⇒i4n−1=1i×ii\Rightarrow {i^{4n - 1}} = \dfrac{1}{i} \times \dfrac{i}{i}
Multiply the terms,
⇒i4n−1=ii2\Rightarrow {i^{4n - 1}} = \dfrac{i}{{{i^2}}}
Substitute the value of i2{i^2},
⇒i4n−1=i−1\Rightarrow {i^{4n - 1}} = \dfrac{i}{{ - 1}}
Simplify the terms,
⇒i4n−1=−i\Rightarrow {i^{4n - 1}} = - i..............….. (2)
Substitute the values from equation (1) and (2) in original expression,
⇒i4n+1−i4n−12=i−(−i)2\Rightarrow \dfrac{{{i^{4n + 1}} - {i^{4n - 1}}}}{2} = \dfrac{{i - \left( { - i} \right)}}{2}
Open the bracket and change the sign accordingly,
⇒i4n+1−i4n−12=i+i2\Rightarrow \dfrac{{{i^{4n + 1}} - {i^{4n - 1}}}}{2} = \dfrac{{i + i}}{2}
Add the term in the numerator,
⇒i4n+1−i4n−12=2i2\Rightarrow \dfrac{{{i^{4n + 1}} - {i^{4n - 1}}}}{2} = \dfrac{{2i}}{2}
Cancel out the common factor,
∴i4n+1−i4n−12=i\therefore \dfrac{{{i^{4n + 1}} - {i^{4n - 1}}}}{2} = i
So, the value of i4n+1−i4n−12\dfrac{{{i^{4n + 1}} - {i^{4n - 1}}}}{2} is ii.

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

Note: The complex numbers are the field C of numbers of the form x+iyx + 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+iyz = x + iy.