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Question: If \(x + iy = \dfrac{{{2^{1008}}}}{{{{\left( {1 + i} \right)}^{2016}}}} + \dfrac{{{{\left( {1 + i} \...

If x+iy=21008(1+i)2016+(1+i)201621008x + iy = \dfrac{{{2^{1008}}}}{{{{\left( {1 + i} \right)}^{2016}}}} + \dfrac{{{{\left( {1 + i} \right)}^{2016}}}}{{{2^{1008}}}}, find xx and yy.

Explanation

Solution

First, take the common power from both numerator and denominator. After that, use the formula (a+b)2=a2+b2+2ab{\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab to find the value of (1+i)2{\left( {1 + i} \right)^2}. Then, substitute the value in the equation and cancel out the common factors. As we know i4=1{i^4} = 1, factorize the power in multiples of 4. Then, substitute the value of i4{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=21008(1+i)2016+(1+i)201621008x + 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=(2(1+i)2)1008+((1+i)22)1008x + 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, (a+b)2=a2+b2+2ab{\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab.
Here, a=1a = 1 and b=ib = i. Then,
(1+i)2=12+i2+2×1×i{\left( {1 + i} \right)^2} = {1^2} + {i^2} + 2 \times 1 \times i
Since i2=1{i^2} = - 1. Substitute the value of i2{i^2} and multiply the terms,
(1+i)2=11+2i{\left( {1 + i} \right)^2} = 1 - 1 + 2i
Subtract the like terms on the right side,
(1+i)2=2i{\left( {1 + i} \right)^2} = 2i
Substitute the value of (1+i)2{\left( {1 + i} \right)^2} in equation (1),
x+iy=(22i)1008+(2i2)1008x + 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=(1i)1008+(i)1008x + iy = {\left( {\dfrac{1}{i}} \right)^{1008}} + {\left( i \right)^{1008}}
As any power of 1 will give 1 as the result.
x+iy=1i1008+i1008x + iy = \dfrac{1}{{{i^{1008}}}} + {i^{1008}}
Since i4=1{i^4} = 1. So, factorize the power in multiples of 4,
x+iy=1(i4)252+(i4)252x + iy = \dfrac{1}{{{{\left( {{i^4}} \right)}^{252}}}} + {\left( {{i^4}} \right)^{252}}
Substitute the value of i4{i^4},
x+iy=1(1)252+(1)252x + iy = \dfrac{1}{{{{\left( 1 \right)}^{252}}}} + {\left( 1 \right)^{252}}
Again, any power of 1 will give 1 as the result.
x+iy=2x + iy = 2
It can be written as,
x+iy=2+0ix + iy = 2 + 0i
Thus, x=2,y=0x = 2,y = 0

Hence, the value of xx is 2, and the value of yy 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 i2=1{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.