Question

The value of ${{(1+i)}^{5}}\times {{(1-i)}^{5}}$ is:
a)-8
b)8i
c)8
d)32

Answer 1

Hint: In this given question, we can use the concept of ${{a}^{5}}\times {{b}^{5}}={{\left( ab \right)}^{5}}$ in order to solve this question. Here, we are given to find the product of ${{(1+i)}^{5}}\text{ }and\text{ }{{(1-i)}^{5}}$. This would become \[{{\left[ (1+i)\times (1-i) \right]}^{5}}\]. Now, using the formula for the product of the sum and difference of two numbers is equal to the difference of their squares.

Complete step-by-step answer:
In this given question, we are asked to find out the value of ${{(1+i)}^{5}}\times {{(1-i)}^{5}}$.
Here, we must know that $i$ corresponds to the value of $\sqrt{-1}$ in order to solve this question and also that ${{a}^{5}}\times {{b}^{5}}={{\left( ab \right)}^{5}}$ to get the required answer. We are going to use the formula for the product of the sum and difference of two numbers is equal to the difference of their squares, that is
$\left( a+b \right)\times \left( a-b \right)=\left( {{a}^{2}}-{{b}^{2}} \right)............(1.1)$
Now, let us start our process of solving.
${{(1+i)}^{5}}\times {{(1-i)}^{5}}$
$={{\left[ (1+i)\times (1-i) \right]}^{5}}$ (as aforesaid)
$={{\left( {{1}^{2}}-{{i}^{2}} \right)}^{5}}$ (using equation 1.1)
\[={{\left( 1-(-1) \right)}^{5}}\text{ }\left( \text{as }{{i}^{2}}\text{=}{{\left( \sqrt{-1} \right)}^{2}}\text{=}-1 \right)\]
$={{\left( 1+1 \right)}^{5}}$
$={{2}^{5}}$
$=32$
Hence, we arrive at our answer as ${{2}^{5}}$ that is equal to 32.
Therefore, the correct option to the given question is option (d) which has the value 32.

Note: In the solution, we used equation (1.1) to simplify the expression and then put the value of ${{i}^{2}}$ to obtain the answer, however, one can expand each term of the product separately by repeated multiplication to obtain the required power of the term. However, the obtained answer will still remain the same and we have to use more steps to arrive at the answer in that way.