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Question: If \[{{\left[ \dfrac{(1+i)}{(1-i)} \right]}^{m}}=1\] then least integral value of \[m\] is 1\. 2 ...

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

Explanation

Solution

To reach the desired result, first simplify the given formula by multiplying (1+i)(1+i)to both numerator and denominator and then substituting i2=1{{i}^{2}}=1 several positive integral values of n starting at 1.

Complete step by step answer:
We are given an expression   [(1+i)(1−i)]m\,\,{{\left[ \dfrac{(1+i)}{(1-i)} \right]}^{m}}, we have to find the least positive integer value of mm, such that the expression should be  1\,1.
Consider the given expression in the question as
 E=[(1+i)(1−i)]m\,E={{\left[ \dfrac{(1+i)}{(1-i)} \right]}^{m}}
First of all, simplify this expression. By multiplying (1+i)(1+i) to both numerator and denominator inside the bracket, we get,
 E=[(1+i)(1+i)(1−i)(1+i)]m\,E={{\left[ \dfrac{(1+i)(1+i)}{(1-i)(1+i)} \right]}^{m}}
We know that(a−b)(a+b)=a2−b2(a-b)(a+b)={{a}^{2}}-{{b}^{2}}. By applying this in the denominator of the above expression, we get,
 E=[(1+i)21−(i)2]m\,E={{\left[ \dfrac{{{(1+i)}^{2}}}{1-{{(i)}^{2}}} \right]}^{m}}
We also know that(a+b)2=a2+2ab+b2{{(a+b)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}. By applying this in numerator of the above expression, we get,
 E=[1+(i)2+2i1−(i)2]m\,E={{\left[ \dfrac{1+{{(i)}^{2}}+2i}{1-{{(i)}^{2}}} \right]}^{m}}
We know that i=−1i=\sqrt{-1}
Bu squaring on both sides, we get
i2=−1{{i}^{2}}=-1
By substituting i2=−1{{i}^{2}}=-1in the above expression, we get,
E=[1−1+2i1−(−1)]mE={{\left[ \dfrac{1-1+2i}{1-(-1)} \right]}^{m}}
We get,  E=[+2i2]m\,E={{\left[ \dfrac{+2i}{2} \right]}^{m}}
Or,  E=(i)m\,E={{(i)}^{m}}
Now, we have to find the least positive integer value of mm such that the above expression should be  1\,1
Substitute m=1m=1 in the above expression
E=(i)1=iE={{(i)}^{1}}=i
But E=iE=i is purely imaginary but not equal to 11 , so m≠1m\ne 1
Substitute m=2m=2 we get:
 E=(i)2=−1\,E={{(i)}^{2}}=-1
But E=−1E=-1 is not equal to 11 , so m≠2m\ne 2
Substitute m=3m=3we get:
 E=(i)3\,E={{(i)}^{3}}
We know that(a)m.(a)m=(a)m+n{{(a)}^{m}}.{{(a)}^{m}}={{(a)}^{m+n}}. By applying this, we get:
E=((i)2(i)1)E=\left( {{(i)}^{2}}{{(i)}^{1}} \right)
We know that i2=−1{{i}^{2}}=-1. Therefore we get,
 E=(−1)i=−i\,E=(-1)i=-i
But E=−iE=-i is purely negative imaginary but not equal to 1\,1 , so m≠3m\ne 3
Substitute m=4m=4 we get:
 E=(i)4\,E={{(i)}^{4}}
We know that(a)m.(a)m=(a)m+n{{(a)}^{m}}.{{(a)}^{m}}={{(a)}^{m+n}}. By applying this, we get:
 E=((i)2.(i)2)\,E=\left( {{(i)}^{2}}.{{(i)}^{2}} \right)
We know that i2=−1{{i}^{2}}=-1Therefore we get,
 E=(−1)(−1)=1\,E=(-1)(-1)=1
But E=1E=1 is exactly equal to 1\,1. Hence, we get m=4m=4
Hence, we get m=4m=4 for which  [(1+i)(1−i)]m=1\,\,{{\left[ \dfrac{(1+i)}{(1-i)} \right]}^{m}}=1.

So, the correct answer is “Option 2”.

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