# Prove the following; If $x - iy = \sqrt {\dfrac{{a - ib}}{{c - id}}} $ prove that ${\left( {{x^2} + {y^2}} \right)^2} = \dfrac{{{a^2} + {b^2}}}{{{c^2} + {d^2}}}$.

Last updated date: 27th Mar 2023

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Hint: The given expression is $x - iy = \sqrt {\dfrac{{a - ib}}{{c - id}}} $, use the concept of taking complex conjugate to this equation keeping one thing in mind that complex conjugate effect only the iota terms and not any real term. This will help you reach the proof in this question.

The given expression is $x - iy = \sqrt {\dfrac{{a - ib}}{{c - id}}} $……………………….. (1)

We need to prove that ${\left( {{x^2} + {y^2}} \right)^2} = \dfrac{{{a^2} + {b^2}}}{{{c^2} + {d^2}}}$……………… (2)

Taking complex conjugate both the sides of equation (1) we get,

$\overline {x - iy} = \overline {\sqrt {\dfrac{{a - ib}}{{c - id}}} } $………………… (3)

Using the property of conjugate $\overline {\left( {a + ib} \right)} = \left( {\overline a + \overline {ib} } \right)$ in equation (3) we get,

$\overline x - \overline {iy} = \sqrt {\dfrac{{\overline a - \overline {ib} }}{{\overline c - \overline {id} }}} $……………… (4)

Now using the property that $\overline x = x{\text{ and }}\overline i = - i$in equation (4) we get,

$x + iy = \sqrt {\dfrac{{a + ib}}{{c + id}}} $……………………… (5)

Now let’s multiply equation (1) and equation (5) we get,

$(x - iy)(x + iy) = \sqrt {\dfrac{{a + ib}}{{c + id}}} \times \sqrt {\dfrac{{a - ib}}{{c - id}}} $

\[ \Rightarrow (x - iy)(x + iy) = \sqrt {\dfrac{{a + ib}}{{c + id}} \times \dfrac{{a - ib}}{{c - id}}} \]……………………….. (6)

Now using the identity $\left( {a - b} \right)\left( {a + b} \right) = {a^2} - {b^2}$ in equation (6) we get,

$ \Rightarrow {x^2} - {i^2}{y^2} = \sqrt {\dfrac{{{a^2} - {i^2}{b^2}}}{{{c^2} - {i^2}{d^2}}}} $

Using ${i^2} = - 1$ we get

$ \Rightarrow {x^2} + {y^2} = \sqrt {\dfrac{{{a^2} + {b^2}}}{{{c^2} + {d^2}}}} $

Squaring both sides,

${\left( {{x^2} + {y^2}} \right)^2} = \dfrac{{{a^2} + {b^2}}}{{{c^2} + {d^2}}}$

Hence proved

Note: Whenever we face such types of problems the key concept is based upon taking complex conjugate and using the various properties of complex conjugate, some of them are being mentioned above while performing the solution. This will help you get on the right track to reach the proof.

The given expression is $x - iy = \sqrt {\dfrac{{a - ib}}{{c - id}}} $……………………….. (1)

We need to prove that ${\left( {{x^2} + {y^2}} \right)^2} = \dfrac{{{a^2} + {b^2}}}{{{c^2} + {d^2}}}$……………… (2)

Taking complex conjugate both the sides of equation (1) we get,

$\overline {x - iy} = \overline {\sqrt {\dfrac{{a - ib}}{{c - id}}} } $………………… (3)

Using the property of conjugate $\overline {\left( {a + ib} \right)} = \left( {\overline a + \overline {ib} } \right)$ in equation (3) we get,

$\overline x - \overline {iy} = \sqrt {\dfrac{{\overline a - \overline {ib} }}{{\overline c - \overline {id} }}} $……………… (4)

Now using the property that $\overline x = x{\text{ and }}\overline i = - i$in equation (4) we get,

$x + iy = \sqrt {\dfrac{{a + ib}}{{c + id}}} $……………………… (5)

Now let’s multiply equation (1) and equation (5) we get,

$(x - iy)(x + iy) = \sqrt {\dfrac{{a + ib}}{{c + id}}} \times \sqrt {\dfrac{{a - ib}}{{c - id}}} $

\[ \Rightarrow (x - iy)(x + iy) = \sqrt {\dfrac{{a + ib}}{{c + id}} \times \dfrac{{a - ib}}{{c - id}}} \]……………………….. (6)

Now using the identity $\left( {a - b} \right)\left( {a + b} \right) = {a^2} - {b^2}$ in equation (6) we get,

$ \Rightarrow {x^2} - {i^2}{y^2} = \sqrt {\dfrac{{{a^2} - {i^2}{b^2}}}{{{c^2} - {i^2}{d^2}}}} $

Using ${i^2} = - 1$ we get

$ \Rightarrow {x^2} + {y^2} = \sqrt {\dfrac{{{a^2} + {b^2}}}{{{c^2} + {d^2}}}} $

Squaring both sides,

${\left( {{x^2} + {y^2}} \right)^2} = \dfrac{{{a^2} + {b^2}}}{{{c^2} + {d^2}}}$

Hence proved

Note: Whenever we face such types of problems the key concept is based upon taking complex conjugate and using the various properties of complex conjugate, some of them are being mentioned above while performing the solution. This will help you get on the right track to reach the proof.

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