Question

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

Answer 1

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: ${{x}^{2}}+1=0$ 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: $\left( 1+i \right)x+\left( 1+i \right)=0$ . $x= -1$ is the root of the equation.
In given question we have two terms:
$S\left( n \right)={{i}^{n}}+{{i}^{-n}}$
First term = ${{i}^{n}}$
Second term = ${{i}^{-n}}$
We know few conditions:
$\begin{align}
  & i=i \\
 & {{i}^{2}}=-1 \\
 & {{i}^{3}}=-i \\
\end{align}$
So by simplifying second term, we get:
${{i}^{-n}}={{\left( {{i}^{-1}} \right)}^{n}}=$ second term
By normal algebraic properties, inverse of a number can be written as:
${{a}^{-1}}=\dfrac{1}{a}$
By using above condition here, we get that:
Second term = $\dfrac{1}{{{i}^{n}}}$
By multiplying and dividing by i inside the power, we get:
Second term = ${{\left( \dfrac{1\times i}{i\times i} \right)}^{n}}$
We know, ${{i}^{2}}=-1$.
By substituting above, we get
Second term = ${{\left( \dfrac{i}{-1} \right)}^{n}}$
Second term = ${{\left( -i \right)}^{n}}$
Given equation in the question is, written as:
$s\left( n \right)={{i}^{n}}+{{i}^{-n}}$
By substituting second term, we get:
$s\left( n \right)={{i}^{n}}+{{\left( -i \right)}^{n}}$
By separating minus sign from second term, we get:
$s\left( n \right)={{i}^{n}}+\left( -1 \right){{i}^{n}}$
By taking common, the common term in first and second terms, we get:
$s\left( n \right)={{i}^{n}}\left( 1+{{\left( -1 \right)}^{n}} \right)$
If in is odd
${{\left( -1 \right)}^{n}}=-1$
Substituting this value in the equation, we get:
$s\left( n \right)={{i}^{n}}\left( 1-1 \right)=0$
If n is even
${{\left( -1 \right)}^{n}}=0$
Substituting this value in the equation, we get:
$s\left( n \right)={{i}^{n}}\left( 1+1 \right)=2{{i}^{n}}$
Now we have 2 cases
Case 1: n is multiple of 4
$s\left( n \right)=2{{i}^{n}}$
Let n = 4k
By substituting n value in equation, we get
$s\left( n \right)=2{{i}^{4k}}$
By writing the equation smartly, we get
$s\left( n \right)=2{{\left( {{i}^{2}} \right)}^{2k}}=2{{\left( -1 \right)}^{2k}}$
${{\left( -1 \right)}^{2k}}=1$
By simplifying, we get
s(n) = 2
Case 2: n is not multiple of 4.
n = 2k where k = odd
$s\left( n \right)=2{{i}^{2k}}=2{{\left( -1 \right)}^{k}}$
as k = odd, we can say:
s(n) = -2
Therefore, the possible distinct values of s(n) are $\{-2,0,+2\}$ .

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.