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).
\[\begin{align}
  & {{S}_{n}}={{i}^{n}}+{{i}^{-n}} \\
 & {{S}_{n}}={{i}^{n}}+{{\left( -i \right)}^{n}} \\
 & {{S}_{n}}={{i}^{n}}+{{\left( \left( -1 \right)\times i \right)}^{n}} \\
\end{align}\]
\[{{S}_{n}}={{i}^{n}}+{{\left( -1 \right)}^{n}}\times {{\left( i \right)}^{n}}\] , 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).