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Question: The expression \({{\left( \dfrac{1+i}{1-i} \right)}^{2}}+{{\left( \dfrac{1-i}{1+i} \right)}^{2}}\) i...

The expression (1+i1−i)2+(1−i1+i)2{{\left( \dfrac{1+i}{1-i} \right)}^{2}}+{{\left( \dfrac{1-i}{1+i} \right)}^{2}} is equal to
a)2i2i
b)−2i-2i
c)−2-2
d)22

Explanation

Solution

Hint: In the given question we can first equalize the LCM of the denominators and multiply the required factors in order to do so in the numerators and denominators of both the terms so that we will be able to add both of these terms. Then we will use the formula for the product of the sum and difference of two numbers is equal to the difference of their squares, that is: (a+b)×(a−b)=(a2−b2)\left( a+b \right)\times \left( a-b \right)=\left( {{a}^{2}}-{{b}^{2}} \right) in the denominator and expansion formula for the whole square of the sum of two numbers that is(a+b)2=a2+b2+2ab{{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab in the numerator and simplify the polynomial to get our answer.

Complete step-by-step answer:
In this given question, we are asked to find out the value of (1+i1−i)2+(1−i1+i)2{{\left( \dfrac{1+i}{1-i} \right)}^{2}}+{{\left( \dfrac{1-i}{1+i} \right)}^{2}}.
Here, we must know that ii corresponds to the value of −1\sqrt{-1} in order to solve this question.
Now, we are going to use the following formulae in order to solve this question and get our required answer:
(a+b)×(a−b)=(a2−b2)............(1.1)\left( a+b \right)\times \left( a-b \right)=\left( {{a}^{2}}-{{b}^{2}} \right)............(1.1)
(a+b)2=a2+b2+2ab............(1.2){{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab............(1.2)
Let us name the given expression as I and start the process of solving the given polynomial and get the required answer as follows:
I=(1+i1−i)2+(1−i1+i)2I={{\left( \dfrac{1+i}{1-i} \right)}^{2}}+{{\left( \dfrac{1-i}{1+i} \right)}^{2}}
=(1+i)2(1−i)2+(1−i)2(1+i)2=\dfrac{{{\left( 1+i \right)}^{2}}}{{{\left( 1-i \right)}^{2}}}+\dfrac{{{\left( 1-i \right)}^{2}}}{{{\left( 1+i \right)}^{2}}}
=(1+i)2×(1+i)2+(1−i)2×(1−i)2(1−i)2×(1+i)2=\dfrac{{{\left( 1+i \right)}^{2}}\times {{\left( 1+i \right)}^{2}}+{{\left( 1-i \right)}^{2}}\times {{\left( 1-i \right)}^{2}}}{{{\left( 1-i \right)}^{2}}\times {{\left( 1+i \right)}^{2}}}
=((1+i)2)2+((1−i)2)2((1−i)×(1+i))2=\dfrac{{{\left( {{\left( 1+i \right)}^{2}} \right)}^{2}}+{{\left( {{\left( 1-i \right)}^{2}} \right)}^{2}}}{{{\left( \left( 1-i \right)\times \left( 1+i \right) \right)}^{2}}}
Using equation 1.1 and 1.2, we can write the above expression as
I=(1+i2+2i)2+(1+i2−2i)2(12−i2)2I=\dfrac{{{\left( 1+{{i}^{2}}+2i \right)}^{2}}+{{\left( 1+{{i}^{2}}-2i \right)}^{2}}}{{{\left( {{1}^{2}}-{{i}^{2}} \right)}^{2}}}
As i=−1i=\sqrt{-1}, this implies that i2=(−1)2=−1{{i}^{2}}={{\left( \sqrt{-1} \right)}^{2}}=-1. Therefore, using the value of i2{{i}^{2}} in the above expression
I=(1+(−1)+2i)2+(1+(−1)−2i)2(12−(−1))2I=\dfrac{{{\left( 1+(-1)+2i \right)}^{2}}+{{\left( 1+(-1)-2i \right)}^{2}}}{{{\left( {{1}^{2}}-(-1) \right)}^{2}}}
=(2i)2+(−2i)222=\dfrac{{{\left( 2i \right)}^{2}}+{{\left( -2i \right)}^{2}}}{{{2}^{2}}}
=4i2+4i24=\dfrac{4{{i}^{2}}+4{{i}^{2}}}{4}
=4(i2+i2)4=\dfrac{4\left( {{i}^{2}}+{{i}^{2}} \right)}{4}
Again using the value of i2{{i}^{2}} to be equal to -1, we get
I=−1+(−1)I=-1+(-1)
=−2=-2
Hence, we get our answer as -2.
Therefore, the correct option to this given question is option (c) which has the value -2.

Note: In the solution, we took the lcm of the denominators of the two terms and then simplified the expression. However, one can simplify each term by multiplying and dividing the first and second term by (1+i)\left( 1+i \right) and (1−i)\left( 1-i \right) respectively. However, the obtained answer would still remain the same.