Answer
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Hint: In the given question, there are two parts. For part one, take the conjugate of the given expression to get the desired result and for the second part multiply the given expression with its conjugate to get the desired result.
Complete step-by-step answer:
It is given that $\dfrac{{a + ib}}{{c + id}} = x + iy$
And we are going to solve this problem in two parts.
In part one we are going to prove $\dfrac{{a - ib}}{{c - id}} = x - iy$
Given expression is $\dfrac{{a + ib}}{{c + id}} = x + iy$, take its conjugate value, because the proof that we are going to prove looks like its conjugate.
And we know one thing that if two complex numbers are equal, then its conjugates must be also equal.
Hence,
\[ \Rightarrow \overline {\left( {\dfrac{{a + ib}}{{c + id}}} \right)} = \overline {\left( {x + iy} \right)} \]
\[ \Rightarrow \left( {\dfrac{{\overline {a + ib} }}{{\overline {c + id} }}} \right) = \left( {\overline {x + iy} } \right)\]
And we know that \[\left( {\overline {x + iy} } \right) = \left( {x - iy} \right)\], then we get
\[ \Rightarrow \left( {\dfrac{{a - ib}}{{c - id}}} \right) = \left( {x - iy} \right)\]
Our first part proof is done.
Let us do the second part,
We need to prove $\dfrac{{{a^2} + {b^2}}}{{{c^2} + {d^2}}} = {x^2} + {y^2}$
For the given expression, $\dfrac{{a + ib}}{{c + id}} = x + iy$ multiply it with its conjugate.
Because the proof has all real numbers.
$ \Rightarrow \left( {\dfrac{{a + ib}}{{c + id}}} \right) \times \left( {\dfrac{{a - ib}}{{c - id}}} \right) = \left( {x + iy} \right) \times \left( {x - iy} \right)$
We also know that $\left( {a + b} \right)\left( {a - b} \right) = \left( {{a^2} - {b^2}} \right)$. Applying this formula in the above equation, we get
$ \Rightarrow \left( {\dfrac{{{a^2} - {{\left( {ib} \right)}^2}}}{{{c^2} - {{\left( {id} \right)}^2}}}} \right) = \left( {{x^2} - {{\left( {iy} \right)}^2}} \right)$
We also know that ${i^2} = - 1$. So,
$ \Rightarrow \left( {\dfrac{{{a^2} + {b^2}}}{{{c^2} + {d^2}}}} \right) = \left( {{x^2} + {y^2}} \right)$
Hence, the second part is also proved.
Note: There is a very good chance of asking this kind of question in the exams. You can do them simply by taking conjugate values or multiplying or dividing them with its conjugates. If the result to be proven is a real number then mostly we need to multiply with conjugate to remove $i$ from the expression.
Conjugate of $a+ib$ is $a-ib$.
Complete step-by-step answer:
It is given that $\dfrac{{a + ib}}{{c + id}} = x + iy$
And we are going to solve this problem in two parts.
In part one we are going to prove $\dfrac{{a - ib}}{{c - id}} = x - iy$
Given expression is $\dfrac{{a + ib}}{{c + id}} = x + iy$, take its conjugate value, because the proof that we are going to prove looks like its conjugate.
And we know one thing that if two complex numbers are equal, then its conjugates must be also equal.
Hence,
\[ \Rightarrow \overline {\left( {\dfrac{{a + ib}}{{c + id}}} \right)} = \overline {\left( {x + iy} \right)} \]
\[ \Rightarrow \left( {\dfrac{{\overline {a + ib} }}{{\overline {c + id} }}} \right) = \left( {\overline {x + iy} } \right)\]
And we know that \[\left( {\overline {x + iy} } \right) = \left( {x - iy} \right)\], then we get
\[ \Rightarrow \left( {\dfrac{{a - ib}}{{c - id}}} \right) = \left( {x - iy} \right)\]
Our first part proof is done.
Let us do the second part,
We need to prove $\dfrac{{{a^2} + {b^2}}}{{{c^2} + {d^2}}} = {x^2} + {y^2}$
For the given expression, $\dfrac{{a + ib}}{{c + id}} = x + iy$ multiply it with its conjugate.
Because the proof has all real numbers.
$ \Rightarrow \left( {\dfrac{{a + ib}}{{c + id}}} \right) \times \left( {\dfrac{{a - ib}}{{c - id}}} \right) = \left( {x + iy} \right) \times \left( {x - iy} \right)$
We also know that $\left( {a + b} \right)\left( {a - b} \right) = \left( {{a^2} - {b^2}} \right)$. Applying this formula in the above equation, we get
$ \Rightarrow \left( {\dfrac{{{a^2} - {{\left( {ib} \right)}^2}}}{{{c^2} - {{\left( {id} \right)}^2}}}} \right) = \left( {{x^2} - {{\left( {iy} \right)}^2}} \right)$
We also know that ${i^2} = - 1$. So,
$ \Rightarrow \left( {\dfrac{{{a^2} + {b^2}}}{{{c^2} + {d^2}}}} \right) = \left( {{x^2} + {y^2}} \right)$
Hence, the second part is also proved.
Note: There is a very good chance of asking this kind of question in the exams. You can do them simply by taking conjugate values or multiplying or dividing them with its conjugates. If the result to be proven is a real number then mostly we need to multiply with conjugate to remove $i$ from the expression.
Conjugate of $a+ib$ is $a-ib$.