Question

Show that the following is purely real
\left\\{ \dfrac{\left( \sqrt{7}+i\sqrt{3} \right)}{\left( \sqrt{7}-i\sqrt{3} \right)}+\dfrac{\left( \sqrt{7}-i\sqrt{3} \right)}{\left( \sqrt{7}+i\sqrt{3} \right)} \right\\}

Answer 1

Hint: First of all, consider the given expression and simplify by taking LCM as $\left( \sqrt{7}-\sqrt{3}i \right)\left( \sqrt{7}+\sqrt{3}i \right)$. By using ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$ and ${{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$, further simplify the expression and substitute ${{i}^{2}}=-1$ in the expression to prove the expression as purely real.

Complete step-by-step answer:
In this question, we have to show that the expression \left\\{ \dfrac{\left( \sqrt{7}+i\sqrt{3} \right)}{\left( \sqrt{7}-i\sqrt{3} \right)}+\dfrac{\left( \sqrt{7}-i\sqrt{3} \right)}{\left( \sqrt{7}+i\sqrt{3} \right)} \right\\} is purely real. Let us consider the expression given in the question.
E=\left\\{ \dfrac{\left( \sqrt{7}+i\sqrt{3} \right)}{\left( \sqrt{7}-i\sqrt{3} \right)}+\dfrac{\left( \sqrt{7}-i\sqrt{3} \right)}{\left( \sqrt{7}+i\sqrt{3} \right)} \right\\}
For the above expression, we have to prove that the given expression is purely real or imaginary part of this expression is 0 which means the term containing i is 0. So by simplifying the above equation, we get,
$E=\dfrac{{{\left( \sqrt{7}+i\sqrt{3} \right)}^{2}}+{{\left( \sqrt{7}-i\sqrt{3} \right)}^{2}}}{\left( \sqrt{7}-i\sqrt{3} \right)\left( \sqrt{7}+i\sqrt{3} \right)}$
We know that, ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$ and ${{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$. By using this in the above expression, we get,
$E=\dfrac{{{\left( \sqrt{7} \right)}^{2}}+{{\left( i\sqrt{3} \right)}^{2}}+2\left( \sqrt{7} \right)\left( i\sqrt{3} \right)+{{\left( \sqrt{7} \right)}^{2}}+{{\left( i\sqrt{3} \right)}^{2}}-2\left( \sqrt{7} \right)\left( i\sqrt{3} \right)}{\left( \sqrt{7}-i\sqrt{3} \right)\left( \sqrt{7}+i\sqrt{3} \right)}$
By simplifying the above equation, we get,
$E=\dfrac{7+3{{i}^{2}}+2\sqrt{21}i+7+3{{i}^{2}}-2\sqrt{21}i}{\left( \sqrt{7}-i\sqrt{3} \right)\left( \sqrt{7}+i\sqrt{3} \right)}$
We know that $\left( a-b \right)\left( a+b \right)={{a}^{2}}-{{b}^{2}}$. By using this in the above equation, we get,
$E=\dfrac{7+3{{i}^{2}}+2\sqrt{21}i+7+3{{i}^{2}}-2\sqrt{21}i}{{{\left( \sqrt{7} \right)}^{2}}-{{\left( i\sqrt{3} \right)}^{2}}}$
$\Rightarrow E=\dfrac{7+3{{i}^{2}}+2\sqrt{21}i+7+3{{i}^{2}}-2\sqrt{21}i}{7-3{{i}^{2}}}$
By canceling the like terms of the above equation, we get,
$E=\dfrac{7+3{{i}^{2}}+7+3{{i}^{2}}}{7-3{{i}^{2}}}$
By rearranging the terms of the above equation, we get,
$E=\dfrac{\left( 7+7 \right)+\left( 3{{i}^{2}}+3{{i}^{2}} \right)}{7-3{{i}^{2}}}$
$E=\dfrac{\left( 14 \right)+\left( 6{{i}^{2}} \right)}{7-3{{i}^{2}}}$
We know that $i=\sqrt{-1}$. By squaring both the sides, we get ${{i}^{2}}=-1$. So by substituting ${{i}^{2}}=-1$ in the above equation, we get,
$E=\dfrac{\left( 14 \right)+6\left( -1 \right)}{7-3\left( -1 \right)}$
$E=\dfrac{14-6}{7+3}$
$E=\dfrac{8}{10}$
By simplifying the above fraction, we get,
$E=\dfrac{4}{5}$
Hence, we have proved that the given expression is purely real, that is the imaginary term or the term containing i is 0.

Note: In these types of questions, students must note that the expression is purely real if it does not contain any term having an imaginary part or i. Similarly, if an expression is purely imaginary if it does not contain any term having real part or it has only terms that contain imaginary part or i. Also, when we say an imaginary number, then it has a real part as well as the imaginary part and is of the form A + iB where A and B are real.