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Question: If root of \((a + ib) = x + iy\), then possible value of root \((a - ib)\) is...

If root of (a+ib)=x+iy(a + ib) = x + iy, then possible value of root (a−ib)(a - ib) is

Explanation

Solution

First, complex numbers are the real and imaginary combined numbers as in the form of z=x+iyz = x + iy, where x and y are the real numbers and ii is the imaginary.
Imaginary ii can be also represented into the real values only if, i2=−1{i^2} = - 1
We may apply the two general formulas in the algebraic form (a+b)2,(a−b)2{(a + b)^2},{(a - b)^2} to solve this problem.
Formula used:
(a+b)2=a2+b2+2ab{(a + b)^2} = {a^2} + {b^2} + 2ab and (a−b)2=a2+b2−2ab{(a - b)^2} = {a^2} + {b^2} - 2ab

Complete step-by-step solution:
Since from the given that we have, root (a+ib)=x+iy(a + ib) = x + iy and which can be written as in the form of (a+ib)=x+iy\sqrt {(a + ib)} = x + iy.
Now squaring on the right-hand side and left-hand side we get, (a+ib)2=(x+iy)2{\sqrt {(a + ib)} ^2} = {(x + iy)^2}
Since the root terms and square terms are cancel each other, then we have a+ib=(x+iy)2a + ib = {(x + iy)^2}
Now apply the first formula in the algebraic form, we get a+ib=(x+iy)2⇒x2+(i)2y2+2xiya + ib = {(x + iy)^2} \Rightarrow {x^2} + {(i)^2}{y^2} + 2xiy
Since the imaginary ii can be also represented into the real values and which is i2=−1{i^2} = - 1. Substituting this value, we get a+ib=x2+(i)2y2+2xiy⇒x2−y2+2xiya + ib = {x^2} + {(i)^2}{y^2} + 2xiy \Rightarrow {x^2} - {y^2} + 2xiy
Hence, we get a+ib=x2−y2+2xiya + ib = {x^2} - {y^2} + 2xiy, since two values are equal.
Now compare the real values and imaginary values separately, we get a=x2−y2,ib=2xiy⇒b=2xya = {x^2} - {y^2},ib = 2xiy \Rightarrow b = 2xy
Thus, the required root value is (a−ib)(a - ib), so let us apply the values in this form, we get (a−ib)=(x2−y2)−2xiy(a - ib) = ({x^2} - {y^2}) - 2xiy
Which is the form of second algebraic form, (a−b)2=a2+b2−2ab{(a - b)^2} = {a^2} + {b^2} - 2ab where i2=−1{i^2} = - 1
Hence, we get (a−ib)=(x2−y2)−2xiy⇒(x−iy)2(a - ib) = ({x^2} - {y^2}) - 2xiy \Rightarrow {(x - iy)^2}
Now taking the square root on both sides we get (a−ib)=(x−iy)2⇒(x−iy)\sqrt {(a - ib)} = \sqrt {{{(x - iy)}^2}} \Rightarrow (x - iy)
Therefore, the possible value of the root (a−ib)(a - ib) is (a−ib)=(x−iy)\sqrt {(a - ib)} = (x - iy).

Note: Since Imaginary ii can be also represented into the real values only if, i2=−1{i^2} = - 1and without this value, we cannot solve the given problem also.
In the perfect square, the roots and square roots are reverse processes, hence they will cancel each other.
The formula of the algebraic form is made by the general definition of multiplication which is (a+b)2=(a+b)(a+b)=a.a+a.b+.ba+b.b=a2+b2+2ab{(a + b)^2} = (a + b)(a + b) = a.a + a.b + .ba + b.b = {a^2} + {b^2} + 2ab.