Solveeit Logo
Login

Question

Question: If \(s(n)={{i}^{n}}+{{i}^{-n}}\) , where \(i=\sqrt{-1}\) and n is an integer, then the total number ...

If s(n)=in+i−ns(n)={{i}^{n}}+{{i}^{-n}} , where i=−1i=\sqrt{-1} and n is an integer, then the total number of distinct values of s(n) is
(a) 1
(b) 2
(c) 3
(d) 4

Explanation

Solution

Hint: Find the conditions possible on n. As given n is an integer the cases possible are n is odd and n is even. So, check the values in both the cases.

Complete step-by-step solution -
The solution of the equation: x2+1=0{{x}^{2}}+1=0 is i. i is an imaginary number. Any number which has an imaginary number in its representation is called a complex number.
An equation containing complex numbers in it, is called a complex equation.
It is possible to have a real root for complex equations.
Example: (1+i)x+(1+i)=0\left( 1+i \right)x+\left( 1+i \right)=0 . x=−1x= -1 is the root of the equation.
In given question we have two terms:
S(n)=in+i−nS\left( n \right)={{i}^{n}}+{{i}^{-n}}
First term = in{{i}^{n}}
Second term = i−n{{i}^{-n}}
We know few conditions:
i=i i2=−1 i3=−i \begin{aligned} & i=i \\\ & {{i}^{2}}=-1 \\\ & {{i}^{3}}=-i \\\ \end{aligned}
So by simplifying second term, we get:
i−n=(i−1)n={{i}^{-n}}={{\left( {{i}^{-1}} \right)}^{n}}= second term
By normal algebraic properties, inverse of a number can be written as:
a−1=1a{{a}^{-1}}=\dfrac{1}{a}
By using above condition here, we get that:
Second term = 1in\dfrac{1}{{{i}^{n}}}
By multiplying and dividing by i inside the power, we get:
Second term = (1×ii×i)n{{\left( \dfrac{1\times i}{i\times i} \right)}^{n}}
We know, i2=−1{{i}^{2}}=-1.
By substituting above, we get
Second term = (i−1)n{{\left( \dfrac{i}{-1} \right)}^{n}}
Second term = (−i)n{{\left( -i \right)}^{n}}
Given equation in the question is, written as:
s(n)=in+i−ns\left( n \right)={{i}^{n}}+{{i}^{-n}}
By substituting second term, we get:
s(n)=in+(−i)ns\left( n \right)={{i}^{n}}+{{\left( -i \right)}^{n}}
By separating minus sign from second term, we get:
s(n)=in+(−1)ins\left( n \right)={{i}^{n}}+\left( -1 \right){{i}^{n}}
By taking common, the common term in first and second terms, we get:
s(n)=in(1+(−1)n)s\left( n \right)={{i}^{n}}\left( 1+{{\left( -1 \right)}^{n}} \right)
If in is odd
(−1)n=−1{{\left( -1 \right)}^{n}}=-1
Substituting this value in the equation, we get:
s(n)=in(1−1)=0s\left( n \right)={{i}^{n}}\left( 1-1 \right)=0
If n is even
(−1)n=0{{\left( -1 \right)}^{n}}=0
Substituting this value in the equation, we get:
s(n)=in(1+1)=2ins\left( n \right)={{i}^{n}}\left( 1+1 \right)=2{{i}^{n}}
Now we have 2 cases
Case 1: n is multiple of 4
s(n)=2ins\left( n \right)=2{{i}^{n}}
Let n = 4k
By substituting n value in equation, we get
s(n)=2i4ks\left( n \right)=2{{i}^{4k}}
By writing the equation smartly, we get
s(n)=2(i2)2k=2(−1)2ks\left( n \right)=2{{\left( {{i}^{2}} \right)}^{2k}}=2{{\left( -1 \right)}^{2k}}
(−1)2k=1{{\left( -1 \right)}^{2k}}=1
By simplifying, we get
s(n) = 2
Case 2: n is not multiple of 4.
n = 2k where k = odd
s(n)=2i2k=2(−1)ks\left( n \right)=2{{i}^{2k}}=2{{\left( -1 \right)}^{k}}
as k = odd, we can say:
s(n) = -2
Therefore, the possible distinct values of s(n) are −2,0,+2\\{-2,0,+2\\} .

Note: While taking multiples of 4 and not multiples of 4 be careful to take k = odd in the latter case. There is a need to keep in mind the concept of exponents and iota. This makes our solution easier.