Question

If $x + iy = \dfrac{{{2^{1008}}}}{{{{\left( {1 + i} \right)}^{2016}}}} + \dfrac{{{{\left( {1 + i} \right)}^{2016}}}}{{{2^{1008}}}}$, find $x$ and $y$.

Answer 1

Hint: First, take the common power from both numerator and denominator. After that, use the formula ${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$ to find the value of ${\left( {1 + i} \right)^2}$. Then, substitute the value in the equation and cancel out the common factors. As we know ${i^4} = 1$, factorize the power in multiples of 4. Then, substitute the value of ${i^4}$. Also, any power of 1 returns 1. After that add the terms to get the final result.

Complete step-by-step answer:
Given: - $x + iy = \dfrac{{{2^{1008}}}}{{{{\left( {1 + i} \right)}^{2016}}}} + \dfrac{{{{\left( {1 + i} \right)}^{2016}}}}{{{2^{1008}}}}$
Take the common power from both numerator and denominator,
$x + iy = {\left( {\dfrac{2}{{{{\left( {1 + i} \right)}^2}}}} \right)^{1008}} + {\left( {\dfrac{{{{\left( {1 + i} \right)}^2}}}{2}} \right)^{1008}}$ ….. (1)
As we know that, ${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$.
Here, $a = 1$ and $b = i$. Then,
${\left( {1 + i} \right)^2} = {1^2} + {i^2} + 2 \times 1 \times i$
Since ${i^2} = - 1$. Substitute the value of ${i^2}$ and multiply the terms,
${\left( {1 + i} \right)^2} = 1 - 1 + 2i$
Subtract the like terms on the right side,
${\left( {1 + i} \right)^2} = 2i$
Substitute the value of ${\left( {1 + i} \right)^2}$ in equation (1),
$x + iy = {\left( {\dfrac{2}{{2i}}} \right)^{1008}} + {\left( {\dfrac{{2i}}{2}} \right)^{1008}}$
Cancel out the common factors from the numerator and denominator,
$x + iy = {\left( {\dfrac{1}{i}} \right)^{1008}} + {\left( i \right)^{1008}}$
As any power of 1 will give 1 as the result.
$x + iy = \dfrac{1}{{{i^{1008}}}} + {i^{1008}}$
Since ${i^4} = 1$. So, factorize the power in multiples of 4,
$x + iy = \dfrac{1}{{{{\left( {{i^4}} \right)}^{252}}}} + {\left( {{i^4}} \right)^{252}}$
Substitute the value of ${i^4}$,
$x + iy = \dfrac{1}{{{{\left( 1 \right)}^{252}}}} + {\left( 1 \right)^{252}}$
Again, any power of 1 will give 1 as the result.
$x + iy = 2$
It can be written as,
$x + iy = 2 + 0i$
Thus, $x = 2,y = 0$

Hence, the value of $x$ is 2, and the value of $y$ is 0.

Note: A complex number is a number that can be expressed in the form a + bi, where a and b are real numbers, and i represents the imaginary unit, satisfying the equation \[{i^2} = - 1\]. Because no real number satisfies this equation, i is called an imaginary number.
Since any part can be 0, so all Real Numbers and Imaginary Numbers are also Complex Numbers.