Question

If $$x + iy = \sqrt {\dfrac{{a + ib}}{{c + id}}} $$, then $${\left( {{x^2} + {y^2}} \right)^2}$$ is equal to
$$\dfrac{{{a^2} + {b^2}}}{{{c^2} + {d^2}}}$$
$$\dfrac{{a + b}}{{c + d}}$$
$$\dfrac{{{c^2} + {d^2}}}{{{a^2} + {b^2}}}$$
$${\left[ {\dfrac{{{a^2} + {b^2}}}{{{c^2} + {d^2}}}} \right]^2}$$

Answer 1

Hint: Here in this question given a complex number, we have to find the value of $${\left( {{x^2} + {y^2}} \right)^2}$$ using a given condition $$x + iy$$. For this first we need to find the conjugate of a given complex number and then multiply both the given complex number and its conjugate and on further simplification we get the required solution.

Complete step-by-step answer:
A complex number generally denoted as Capital Z $$\left( Z \right)$$ is any number that can be written in the form $$x + iy$$ it’s always represented in binomial form. Where, $$x$$ and $$y$$ are real numbers. ‘$$x$$’ is called the real part of the complex number, ‘$$y$$’ is called the imaginary part of the complex number, and ‘$$i$$’ (iota) is called the imaginary unit.
Consider, the given complex number
$$x + iy = \sqrt {\dfrac{{a + ib}}{{c + id}}} $$ -----(1)
We need to find the value of $${\left( {{x^2} + {y^2}} \right)^2}$$
First, we need to find conjugate of (1) i.e.,
$$\,\,x - iy = \sqrt {\dfrac{{a - ib}}{{c - id}}} $$ ------(2)
Multiplying equation (1) by (2), then we get
$$ \Rightarrow \,\,\left( {x + iy} \right)\left( {x - iy} \right) = \left( {\sqrt {\dfrac{{a + ib}}{{c + id}}} } \right)\left( {\sqrt {\dfrac{{a - ib}}{{c - id}}} } \right)$$
We know that $$\sqrt a \sqrt b = \sqrt {ab} $$, then we have
$$ \Rightarrow \,\,\left( {x + iy} \right)\left( {x - iy} \right) = \left( {\sqrt {\dfrac{{\left( {a + ib} \right)\left( {a - ib} \right)}}{{\left( {c + id} \right)\left( {c - id} \right)}}} } \right)$$
Now apply a algebraic identity $${a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)$$ in both LHS and RHS, then we have
$$ \Rightarrow \,\,{x^2} - {\left( {iy} \right)^2} = \left( {\sqrt {\dfrac{{{a^2} - {{\left( {ib} \right)}^2}}}{{{c^2} - {{\left( {id} \right)}^2}}}} } \right)$$
As we know the value of $$i = \sqrt { - 1} $$ and $${i^2} = - 1$$, then
$$ \Rightarrow \,\,{x^2} + {y^2} = \sqrt {\dfrac{{{a^2} + {b^2}}}{{{c^2} + {d^2}}}} $$
Squaring on both the sides, we get
$$ \Rightarrow \,\,{\left( {{x^2} + {y^2}} \right)^2} = {\left( {\sqrt {\dfrac{{{a^2} + {b^2}}}{{{c^2} + {d^2}}}} } \right)^2}$$
On simplification, we get
$$\therefore \,\,{\left( {{x^2} + {y^2}} \right)^2} = \dfrac{{{a^2} + {b^2}}}{{{c^2} + {d^2}}}$$
Hence, it’s a required solution.
Therefore, option (A) is the correct answer.
So, the correct answer is “Option A”.

Note: Always complex numbers are in the form of $$Z = x + iy$$ where $$i$$ is an imaginary number whose value is $$i = \sqrt { - 1} $$. Remember the conjugate of a complex number represented as ‘Z bar’ ($$\overline Z $$) is the number with an equal real part and an imaginary part equal in magnitude but opposite in sign and must know the basic algebraic identities.