Question

The value of $ {\left( {\dfrac{{1 + i}}{{1 - i}}} \right)^2} + {\left( {\dfrac{{1 - i}}{{1 + i}}} \right)^2} $ is equal to:
A. 2i
B. – 2i
C. – 2
D. 2

Answer 1

Hint: Here, Consider the two given terms of the expression separately and simplify each term one by one. To simplify the terms rationalize the denominator to get the term in the form of a + ib. At the last add the two simplified terms obtained to get the final result. Adding complex numbers can be real numbers or imaginary numbers or complex numbers, so do the calculations carefully.

Complete step-by-step answer:
 Given expression is $ {\left( {\dfrac{{1 + i}}{{1 - i}}} \right)^2} + {\left( {\dfrac{{1 - i}}{{1 + i}}} \right)^2} $
Consider $ {\left( {\dfrac{{1 + i}}{{1 - i}}} \right)^2} $
Rationalizing the denominator, we get
  $ \Rightarrow {\left( {\dfrac{{1 + i}}{{1 - i}} \times \dfrac{{1 + i}}{{1 + i}}} \right)^2} = {\left( {\dfrac{{{{\left( {1 + i} \right)}^2}}}{{{1^2} - {i^2}}}} \right)^2} $
  $ = {\left( {\dfrac{{1 + {i^2} + 2i}}{{1 + 1}}} \right)^2} = {\left( {\dfrac{{1 - 1 + 2i}}{2}} \right)^2} = {i^2} = - 1 $ $ \left[ {{i^2} = - 1} \right] $
Consider $ {\left( {\dfrac{{1 - i}}{{1 + i}}} \right)^2} $
Rationalizing the denominator, we get
  $ \Rightarrow {\left( {\dfrac{{1 - i}}{{1 + i}} \times \dfrac{{1 - i}}{{1 - i}}} \right)^2} = {\left( {\dfrac{{{{\left( {1 - i} \right)}^2}}}{{{1^2} - {i^2}}}} \right)^2} $
  $ = {\left( {\dfrac{{1 + {i^2} - 2i}}{{1 + 1}}} \right)^2} = {\left( {\dfrac{{1 - 1 - 2i}}{2}} \right)^2} = {( - i)^2} = {i^2} = - 1 $
So, $ {\left( {\dfrac{{1 + i}}{{1 - i}}} \right)^2} + {\left( {\dfrac{{1 - i}}{{1 + i}}} \right)^2} $ = − 1 + (− 1) = − 1 – 1 = − 2

So, the correct answer is “Option C”.

Note: In these types of questions, where expression is given in complex number form; simplify the terms using appropriate method and rules. Complex numbers are different from real numbers as complex numbers do not exist and cannot be mentioned on real axis graph paper. In rationalization of a complex number also try that no complex number is there in the denominator. Standard and simplified form of complex number is considered in the form (a + ib), where a is real part and ib is imaginary part although b is a real number. In two complex numbers, real part and imaginary part cannot be added, here two real parts and two imaginary parts can be added only.
All algebraic identities are applicable for complex numbers just remember that $ \left[ {{i^2} = - 1} \right] $ , and put the value in the final result whenever the result is found in square form of i.