Question

If $ {\left( {\dfrac{{1 + i}}{{1 - i}}} \right)^m} = 1, $ the least positive integral value of $ m $ is…

Answer 1

Hint: As we know that the above question consists of complex numbers. A complex number is of the form of $ a + ib $ , where $ a,b $ are real numbers and $ i $ represents the imaginary unit. The term $ i $ is an imaginary number and it is called iota and it has the value of $ \sqrt { - 1} $ . These numbers are not really imaginary. In this question we will first rationalize the denominator and then by using algebraic and exponential, formulas we will solve it.

Complete step by step solution:
As per the given question we have $ {\left( {\dfrac{{1 + i}}{{1 - i}}} \right)^m} = 1 $ .
So we will first take the left hand side and rationalize the denominator by multiplying it with the opposite sign number i.e.
 $ \Rightarrow {\left( {\dfrac{{1 + i}}{{1 - i}} \times \dfrac{{1 + i}}{{1 + i}}} \right)^m} = 1 $ .
We can write it as
 $ \Rightarrow {\left( {\dfrac{{{{(1 + i)}^2}}}{{(1 + i)(1 - i)}}} \right)^m} = 1 $ ,
Here we can see that the denominator is in the form of $ (a + b)(a - b) = {a^2} - {b^2} $ ,
 And also there is another algebraic formula for the numerator which is $ {(a + b)^2} = {a^2} + 2ab + {b^2} $ .
 BY applying both the formulas in the numerator and the denominator we get:
 $ $ $ {\left( {\dfrac{{{1^2} + 2 \times 1 \times i + {i^2}}}{{{1^2} - {i^2}}}} \right)^m} = 1 $ .
 We know that $ {i^2} = - 1 $ , by substituting this we get:
 $ \Rightarrow{\left( {\dfrac{{1 + 2i - 1}}{{1 - ( - 1)}}} \right)^m} = 1 \Rightarrow {\left( {\dfrac{{2i}}{2}} \right)^m} = 1 $ ,
 Therefore it gives us $ {i^m} = 1 $ .
Here we will use the basic exponential rule, we know that $ {i^4} $ gives the value $ 1 $ , so we can write: $ {i^m} = {1^4} $ .
Hence the required value of $ m $ is $ 4 $ .
So, the correct answer is “4”.

Note: We should be careful while solving this kind of questions and we need to be aware of the matrices, their properties and the exponential formulas. We should note that the value of ${i^2}$ is mistakenly taken as $1$, which is a completely wrong value and it may lead to the wrong answers. We should be careful with complex numbers. So ${i^4} = \sqrt { - 1} \times \sqrt { - 1} \times \sqrt { - 1} \times \sqrt { - 1} = 1$.