Question

If \[{{\left[ \dfrac{(1+i)}{(1-i)} \right]}^{m}}=1\] then least integral value of \[m\] is
1. 2
2. 4
3. 5
4. None of these

Answer 1

Hint: To reach the desired result, first simplify the given formula by multiplying \[(1+i)\]to both numerator and denominator and then substituting \[{{i}^{2}}=1\] several positive integral values of n starting at 1.

Complete step by step answer:
We are given an expression \[\,\,{{\left[ \dfrac{(1+i)}{(1-i)} \right]}^{m}}\], we have to find the least positive integer value of \[m\], such that the expression should be \[\,1\].
Consider the given expression in the question as
\[\,E={{\left[ \dfrac{(1+i)}{(1-i)} \right]}^{m}}\]
First of all, simplify this expression. By multiplying \[(1+i)\] to both numerator and denominator inside the bracket, we get,
\[\,E={{\left[ \dfrac{(1+i)(1+i)}{(1-i)(1+i)} \right]}^{m}}\]
We know that\[(a-b)(a+b)={{a}^{2}}-{{b}^{2}}\]. By applying this in the denominator of the above expression, we get,
\[\,E={{\left[ \dfrac{{{(1+i)}^{2}}}{1-{{(i)}^{2}}} \right]}^{m}}\]
We also know that\[{{(a+b)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}\]. By applying this in numerator of the above expression, we get,
\[\,E={{\left[ \dfrac{1+{{(i)}^{2}}+2i}{1-{{(i)}^{2}}} \right]}^{m}}\]
We know that \[i=\sqrt{-1}\]
Bu squaring on both sides, we get
\[{{i}^{2}}=-1\]
By substituting \[{{i}^{2}}=-1\]in the above expression, we get,
\[E={{\left[ \dfrac{1-1+2i}{1-(-1)} \right]}^{m}}\]
We get, \[\,E={{\left[ \dfrac{+2i}{2} \right]}^{m}}\]
Or, \[\,E={{(i)}^{m}}\]
Now, we have to find the least positive integer value of \[m\] such that the above expression should be \[\,1\]
Substitute \[m=1\] in the above expression
\[E={{(i)}^{1}}=i\]
But \[E=i\] is purely imaginary but not equal to \[1\] , so \[m\ne 1\]
Substitute \[m=2\] we get:
\[\,E={{(i)}^{2}}=-1\]
But \[E=-1\] is not equal to \[1\] , so \[m\ne 2\]
Substitute \[m=3\]we get:
\[\,E={{(i)}^{3}}\]
We know that\[{{(a)}^{m}}.{{(a)}^{m}}={{(a)}^{m+n}}\]. By applying this, we get:
\[E=\left( {{(i)}^{2}}{{(i)}^{1}} \right)\]
We know that \[{{i}^{2}}=-1\]. Therefore we get,
\[\,E=(-1)i=-i\]
But \[E=-i\] is purely negative imaginary but not equal to\[\,1\] , so \[m\ne 3\]
Substitute \[m=4\] we get:
\[\,E={{(i)}^{4}}\]
We know that\[{{(a)}^{m}}.{{(a)}^{m}}={{(a)}^{m+n}}\]. By applying this, we get:
\[\,E=\left( {{(i)}^{2}}.{{(i)}^{2}} \right)\]
We know that \[{{i}^{2}}=-1\]Therefore we get,
\[\,E=(-1)(-1)=1\]
But \[E=1\] is exactly equal to\[\,1\]. Hence, we get \[m=4\]
Hence, we get \[m=4\] for which\[\,\,{{\left[ \dfrac{(1+i)}{(1-i)} \right]}^{m}}=1\].

So, the correct answer is “Option 2”.

Note: Some students start substituting \[m=1,2,3,4\] in the original expression that is in \[\,\,{{\left[ \dfrac{(1+i)}{(1-i)} \right]}^{m}}\] only. However, it is preferable to first simplify the formula before substituting the values of \[\,m\] as in the preceding situation, the question becomes somewhat lengthy. Also, after you've given \[\,E={{(i)}^{m}}\], Students frequently make errors and obtain the answer \[m=1\], but they must remember that the original formula should be 1.