Answer

Verified

447.9k+ views

**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 $a-ib$.

Recently Updated Pages

How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE

Mark and label the given geoinformation on the outline class 11 social science CBSE

When people say No pun intended what does that mea class 8 english CBSE

Name the states which share their boundary with Indias class 9 social science CBSE

Give an account of the Northern Plains of India class 9 social science CBSE

Change the following sentences into negative and interrogative class 10 english CBSE

Trending doubts

Difference Between Plant Cell and Animal Cell

Which are the Top 10 Largest Countries of the World?

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

How do you graph the function fx 4x class 9 maths CBSE

Change the following sentences into negative and interrogative class 10 english CBSE

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