Question
Question: If \(x + iy = \dfrac{{{2^{1008}}}}{{{{\left( {1 + i} \right)}^{2016}}}} + \dfrac{{{{\left( {1 + i} \...
If x+iy=(1+i)201621008+21008(1+i)2016, find x and y.
Solution
First, take the common power from both numerator and denominator. After that, use the formula (a+b)2=a2+b2+2ab to find the value of (1+i)2. Then, substitute the value in the equation and cancel out the common factors. As we know i4=1, factorize the power in multiples of 4. Then, substitute the value of i4. 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=(1+i)201621008+21008(1+i)2016
Take the common power from both numerator and denominator,
x+iy=((1+i)22)1008+(2(1+i)2)1008 ….. (1)
As we know that, (a+b)2=a2+b2+2ab.
Here, a=1 and b=i. Then,
(1+i)2=12+i2+2×1×i
Since i2=−1. Substitute the value of i2 and multiply the terms,
(1+i)2=1−1+2i
Subtract the like terms on the right side,
(1+i)2=2i
Substitute the value of (1+i)2 in equation (1),
x+iy=(2i2)1008+(22i)1008
Cancel out the common factors from the numerator and denominator,
x+iy=(i1)1008+(i)1008
As any power of 1 will give 1 as the result.
x+iy=i10081+i1008
Since i4=1. So, factorize the power in multiples of 4,
x+iy=(i4)2521+(i4)252
Substitute the value of i4,
x+iy=(1)2521+(1)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 i2=−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.