Question

Show that $\left( {\dfrac{{\sqrt 7 + i\sqrt 3 }}{{\sqrt 7 - i\sqrt 3 }} + \dfrac{{\sqrt 7 - i\sqrt 3 }}{{\sqrt 7 + i\sqrt 3 }}} \right)$ is real.

Answer 1

This is a problem based on the complex numbers. Here we just need to add these two complex numbers. But before that we just need to rationalise both the numbers individually. Then we will add and will see the final answer whether it is real or imaginary.

Complete step-by-step answer:
First, we assume that given question to $X + Y$
$X$ means,
$X = \dfrac{{\sqrt 7 + i\sqrt 3 }}{{\sqrt 7 - i\sqrt 3 }}$ and
$Y$ means,
$Y = \dfrac{{\sqrt 7 - i\sqrt 3 }}{{\sqrt 7 + i\sqrt 3 }}$
We find that $X + Y$ is real.
First, we find proper form of $X$
$X = \dfrac{{\sqrt 7 + i\sqrt 3 }}{{\sqrt 7 - i\sqrt 3 }}$
We used to rationalize conjugate method,
Conjugate means multiple the given number numerator or denominator opposite the symbol.it is a conjugate method.
Here given number is
$X = \dfrac{{\sqrt 7 + i\sqrt 3 }}{{\sqrt 7 - i\sqrt 3 }}$
So we used to conjugate method here
$X = \dfrac{{\sqrt 7 + i\sqrt 3 }}{{\sqrt 7 - i\sqrt 3 }} \times \dfrac{{\sqrt 7 + i\sqrt 3 }}{{\sqrt 7 + i\sqrt 3 }}$
In this number we used to multiply the conjugate of denominator value.
Multiply on numerator to numerator and denominator to the denominator.
$X = \dfrac{{{{\left( {\sqrt 7 + i\sqrt 3 } \right)}^2}}}{{{{\left( {\sqrt 7 } \right)}^2} - {{\left( {i\sqrt 3 } \right)}^2}}}$
Here we used some formulae
Those formulae are

$${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2} \\\ (a + b)(a - b) = {a^2} - {b^2} \\\$$

So that,
${\left( {\sqrt 7 + i\sqrt 3 } \right)^2} = {\left( {\sqrt 7 } \right)^2} + 2i\sqrt 7 \sqrt 3 + {\left( {i\sqrt 3 } \right)^2}$
Putting ${i^2} = - 1$
$= 7 + 2i\sqrt {21} - 3$
$= 4 + 2i\sqrt {21}$
Then

$${\left( {\sqrt 7 } \right)^2} - {\left( {i\sqrt 3 } \right)^2} = 7 + 3 \\\ = 10 \\\$$

And we apply to $X$
$X = \dfrac{{4 + 2i\sqrt {21} }}{{10}}$
And now we find $Y$ values
$Y = \dfrac{{\sqrt 7 - i\sqrt 3 }}{{\sqrt 7 + i\sqrt 3 }}$
So we used to rationalizing conjugate method here
$Y = \dfrac{{\sqrt 7 - i\sqrt 3 }}{{\sqrt 7 + i\sqrt 3 }} \times \dfrac{{\sqrt 7 - i\sqrt 3 }}{{\sqrt 7 - i\sqrt 3 }}$
In this number we used to multiply the conjugate of denominator value.
Multiply on numerator to numerator and denominator to the denominator.
$X = \dfrac{{{{\left( {\sqrt 7 - i\sqrt 3 } \right)}^2}}}{{{{\left( {\sqrt 7 } \right)}^2} - {{\left( {i\sqrt 3 } \right)}^2}}}$
So that,

$${\left( {\sqrt 7 - i\sqrt 3 } \right)^2} = {\left( {\sqrt 7 } \right)^2} - 2i\sqrt 7 \sqrt 3 + {\left( {i\sqrt 3 } \right)^2} \\\ = 7 - 2i\sqrt {21} - 3 \\\ = 4 - 2i\sqrt {21} \\\$$

Previous terms used formulae are ${\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}$
Then

$${\left( {\sqrt 7 } \right)^2} - {\left( {i\sqrt 3 } \right)^2} = 7 + 3 \\\ = 10 \\\$$

And we apply to $Y$
$Y = \dfrac{{4 - 2i\sqrt {21} }}{{10}}$
Now we add to $X$and $Y$ values

$$X + Y = \dfrac{{4 + 2i\sqrt {21} }}{{10}} + \dfrac{{4 - 2i\sqrt {21} }}{{10}} \\\ = \dfrac{{4 + 2i\sqrt {21} + 4 - 2i\sqrt {21} }}{{10}} \\\$$

The denominator is the same on $X$and $Y$
So merge to two terms.
now add the values,

$$= \dfrac{8}{{10}} \\\ = \dfrac{4}{5} \\\$$

Now we get $\dfrac{4}{5}$
This is not a complex value. because this value does not include $i$ the term.
So this is real. Finally, we get the real value of the given question.

Note: Any number which is in the form of $a+ib$ is called a Complex number. where $i = \sqrt{-1}$. Here a is the real part and b is the imaginary part. If a number is not in the form of a+ib is known as a real number.