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.