Question

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

Answer 1

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

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

Note: Since Imaginary $i$ can be also represented into the real values only if, ${i^2} = - 1$and 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 ${(a + b)^2} = (a + b)(a + b) = a.a + a.b + .ba + b.b = {a^2} + {b^2} + 2ab$.