Question

If $x + iy = \sqrt {\dfrac{{a + ib}}{{c + id}}} $ , prove that ${\left( {{x^2} + {y^2}} \right)^2} = \dfrac{{{a^2} + {b^2}}}{{{c^2} + {d^2}}}$.

Answer 1

Hint:To solve this problem we will find the conjugate of the given complex number. The conjugate of the complex number can be defined as the number which has identical real parts but has imaginary parts which are equal in magnitude but opposite in sign. Further we will use the identity to resolve the expression. We will substitute the square of iota as $ - 1$ to get the final answer.

Complete answer:
The given expression is,
 $x + iy = \sqrt {\dfrac{{a + ib}}{{c + id}}} $ ……(i)

Now we will find the conjugate of $x + iy$ by changing the sign of the imaginary part as in the conjugate real part remains the same but the imaginary part remains the same in magnitude but opposite in sign. Hence conjugate of $x + iy$ is expressed as,
$\begin{array}{l}
\overline {x + iy} = \sqrt {\dfrac{{\overline {a + ib} }}{{\overline {c + id} }}} \\
x - iy = \sqrt {\dfrac{{a - ib}}{{c - id}}} \;
\end{array}$……(ii)
Now we will find the product of $x + iy$ and of $x - iy$ from equation (i) and (ii) . The product can be expressed as:
\[\left( {x + iy} \right)\left( {x - iy} \right) = \sqrt {\dfrac{{a + ib}}{{c + id}}} \; \times \sqrt {\dfrac{{a - ib}}{{c - id}}} \]
On further resolving the above expression we get,
\[{x^2} - ixy + ixy - \left( {{i^2}{y^2}} \right) = \sqrt {\dfrac{{\left( {a + ib} \right)\left( {a - ib} \right)}}{{\left( {c + id} \right)\left( {c - id} \right)}}} \]
On substituting $ - 1$ for ${i^2}$ we will get,
 \[{x^2} + {y^2} = \sqrt {\dfrac{{{a^2} - iab + iab - \left( {{i^2}{b^2}} \right)}}{{{c^2} - icd + icd - \left( {{i^2}{d^2}} \right)}}} \;\]
Again substitute $ - 1$ for ${i^2}$, so the above expression will becomes,
\[{x^2} + {y^2} = \sqrt {\dfrac{{{a^2} + {b^2}}}{{{c^2} + {d^2}}}} \]
On squaring the above equation on both the sides, we get
\[{\left( {{x^2} + {y^2}} \right)^2} = \dfrac{{{a^2} + {b^2}}}{{{c^2} + {d^2}}}\]
Hence, it is proved that left hand side is equal to right hand side that is \[{\left( {{x^2} + {y^2}} \right)^2} = \dfrac{{{a^2} + {b^2}}}{{{c^2} + {d^2}}}\].

Note:The value of the square of iota is $ - 1$ therefore while solving the different expression take care of the sign changing during the calculation. Also take care of the signs while opening the brackets. The above problem can also be solved by assigning a variable $z$ to the value $x + iy$ and finding its conjugate. Both the methods are equally lengthy.