
Prove the following; If $x - iy = \sqrt {\dfrac{{a - ib}}{{c - id}}} $ prove that ${\left( {{x^2} + {y^2}} \right)^2} = \dfrac{{{a^2} + {b^2}}}{{{c^2} + {d^2}}}$.
Answer
607.2k+ views
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.
Recently Updated Pages
Two men on either side of the cliff 90m height observe class 10 maths CBSE

What happens to glucose which enters nephron along class 10 biology CBSE

Cutting of the Chinese melon means A The business and class 10 social science CBSE

Write a dialogue with at least ten utterances between class 10 english CBSE

Show an aquatic food chain using the following organisms class 10 biology CBSE

A circle is inscribed in an equilateral triangle and class 10 maths CBSE

Trending doubts
Why is there a time difference of about 5 hours between class 10 social science CBSE

Write a letter to the principal requesting him to grant class 10 english CBSE

What is the median of the first 10 natural numbers class 10 maths CBSE

The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths

Which of the following does not have a fundamental class 10 physics CBSE

State and prove converse of BPT Basic Proportionality class 10 maths CBSE

