Question

If $S\left( n \right)={{i}^{n}}+{{i}^{-n}}$, 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 value of ${{i}^{-n}}$, multiply (i) to it and get the value. Substitute in S (n) and simplify it. Take the care when n is odd and even. In case of even finding where 4 is a multiple and not a multiple of n. Thus find the values of S (n).

Complete step-by-step answer:
We have been given the expression, $S\left( n \right)={{i}^{n}}+{{i}^{-n}}-(1)$
We know the basics of complex number’s as,
$i=i,{{i}^{2}}=-1,{{i}^{3}}=-1$ and ${{i}^{4}}=-1$.
Now, ${{i}^{-n}}$ can be written as $\dfrac{1}{{{i}^{n}}}$.
${{i}^{n}}=\dfrac{1}{{{i}^{n}}}={{\left( \dfrac{1}{i} \right)}^{n}}$
Now in the above expression let us multiply by ‘i’ in the numerator and denominator.
$\therefore {{\left( \dfrac{1}{i} \right)}^{n}}={{\left( \dfrac{1\times i}{i\times i} \right)}^{n}}={{\left( \dfrac{i}{{{i}^{2}}} \right)}^{n}}$
We discussed above that, $i=i$ and ${{i}^{2}}=-1$.
Thus, ${{\left( \dfrac{i}{{{i}^{2}}} \right)}^{n}}={{\left( \dfrac{i}{-1} \right)}^{n}}$
i.e. $\dfrac{1}{{{i}^{n}}}={{\left( \dfrac{i}{-1} \right)}^{n}}={{\left( -i \right)}^{n}}$
Thus, we got the value of ${{i}^{-n}}$ as ${{\left( -i \right)}^{n}}$.
${{i}^{-n}}={{\left( -i \right)}^{n}}$
Now let us substitute the value of ${{i}^{-n}}$ in (1).

& {{S}_{n}}={{i}^{n}}+{{i}^{-n}} \\\ & {{S}_{n}}={{i}^{n}}+{{\left( -i \right)}^{n}} \\\ & {{S}_{n}}={{i}^{n}}+{{\left( \left( -1 \right)\times i \right)}^{n}} \\\ \end{aligned}$$ $${{S}_{n}}={{i}^{n}}+{{\left( -1 \right)}^{n}}\times {{\left( i \right)}^{n}}$$ , Now take $${{i}^{n}}$$ common. $${{S}_{n}}={{i}^{n}}\left[ 1+{{\left( -1 \right)}^{n}} \right]$$ If n is odd, then $${{S}_{n}}=0$$. For example if n = 3, $${{S}_{n}}={{i}^{3}}\left[ 1+{{\left( -1 \right)}^{3}} \right]$$ $${{S}_{n}}={{i}^{3}}\left[ 1-1 \right]={{i}^{3}}\times 0=0$$ Thus for n = 3, we got $${{S}_{n}}=0$$. Thus when n is odd, $${{S}_{n}}=0$$. If n is even, then let's find out what happens. Let us consider n as a multiple of 4, because $${{i}^{4}}=1$$, thus to get an even positive value we need to consider n as 4. $$\therefore {{S}_{n}}={{i}^{4}}\left[ 1+{{\left( -1 \right)}^{4}} \right]=1\left[ 1+1 \right]=2$$ Now if n is not a multiple of 4, let us take 6. $$\therefore {{S}_{n}}={{i}^{6}}\left[ 1+{{\left( -1 \right)}^{6}} \right]=-1\left( 1+1 \right)=-1\times 2=-2$$ Thus, for even we got 2 distinct values 2 and -2. Thus we have a total of 3 distinct values of $${{S}_{n}}=-2,0,2$$. $$\therefore $$ Option (c) is the correct answer. Note: It is important that you simplify and get the value of $${{i}^{-n}}$$. It is said that n is an integer. It can be odd or even positive or negative. Thus remember to find the cases when n is odd and even to get the distinct values of S (n).