QUESTION

# Let ${{\text{z}}_1} = \dfrac{{2\sqrt 3 + i6\sqrt 7 }}{{6\sqrt 7 + i2\sqrt 3 }}$ and ${{\text{z}}_2} = \dfrac{{\sqrt {11} + i3\sqrt {13} }}{{3\sqrt {13} + i\sqrt {11} }}$, then, $\left| {\dfrac{1}{{{z_1}}} + \dfrac{1}{{{z_2}}}} \right|$ is equal to$(a) 47$$(b) 264$(c){\text{ }}\left| {{z_1} - {z_2}} \right| \\$(d){\text{ }}\left| {{z_1} + {z_2}} \right| \\$$(e){\text{ }}\left| {{z_1}{z_2}} \right| \\$

Hint: Use rationalisation to simplify the expression and use the properties of complex numbers for further solving.

Rationalising $z_1$,

${{\text{z}}_1} = \dfrac{{2\sqrt 3 + i6\sqrt 7 }}{{6\sqrt 7 + i2\sqrt 3 }} \\$

We get,

${{\text{z}}_1} = \dfrac{{2\sqrt 3 + i6\sqrt 7 }}{{6\sqrt 7 + i2\sqrt 3 }} \times \dfrac{{6\sqrt 7 - i2\sqrt 3 }}{{6\sqrt 7 - i2\sqrt 3 }} \\$

$= \dfrac{{12\sqrt {21} + 12\sqrt {21} + 252i - 12i}}{{252 + 12}} \\$

$= \dfrac{{24\sqrt {21} + 240i}}{{264}}{\text{ = }}\dfrac{{\sqrt {21} }}{{11}} + \dfrac{{10i}}{{11}} \\$

Now,

$\dfrac{{\sqrt {21} }}{{11}} - \dfrac{{10i}}{{11}} = {\text{ }}\overline {{{\text{z}}_1}} \\$

Now, rationalising $z_2$,

${{\text{z}}_2} = \dfrac{{\sqrt {11} + i3\sqrt{13}}}{{3\sqrt{13} - i\sqrt {11} }} \\$

We get,

${{\text{z}}_2} = \dfrac{{\sqrt {11} + i3\sqrt{13} }}{{3\sqrt{13} - i\sqrt {11} }} \times \dfrac{{3\sqrt{13} + i\sqrt {11} }}{{3\sqrt{13} + i\sqrt {11} }}{\text{ }} \\$

$= \dfrac{{3\sqrt {143} - 3\sqrt {143} + 117i + 11i}}{{117 + 11}} \\$

$= \dfrac{{128i}}{{128}} = i \\$

$\Rightarrow -i = \overline {{{\text{z}}_2}} \\$

We know that, $\dfrac{1}{{{z_1}}} = \dfrac{{\overline {{z_1}} }}{{{{\left| {{z_1}} \right|}^2}}} \\$

Hence, $\dfrac{1}{{{z_1}}} = \dfrac{{\dfrac{{\sqrt {21} }}{{11}} - \dfrac{{10i}}{{11}}}}{\Bigg({\sqrt {\dfrac{{21}}{{121}} + \dfrac{{100}}{{121}}}\Bigg)^2 }}{\text{ = }}\dfrac{{\sqrt {21} }}{{11}} - \dfrac{{10i}}{{11}} = {\text{ }}\overline {{{\text{z}}_1}} {\text{ }} \\$

Similarly, $\dfrac{1}{{{z_2}}}{\text{ = }}\dfrac{{\overline {{z_2}} }}{{{{\left| {{z_2}} \right|}^2}}}{\text{ = }}\dfrac{-i}{{ 1 }} = -i = {\text{ }}\overline {{{\text{z}}_2}} \\$

$\therefore \; \dfrac{1}{{{z_1}}} + \dfrac{1}{{{z_2}}} = \overline {{{\text{z}}_1}} + \overline {{{\text{z}}_2}} = \overline {{z_1} + {z_2}} \\$

Hence, ${\text{ }}\left| {\dfrac{1}{{{z_1}}} + \dfrac{1}{{{z_2}}}} \right|{\text{ = }}\left| {\overline {{z_1} + {z_2}} } \right| = \left| {{{\text{z}}_1} + {z_2}} \right|\;\;\;\;(\because \left| {\overline z } \right|{\text{ = }}\left| z \right|) \\$

So, option d is the right answer.

Note: Whenever we encounter such a problem, we simply need to use rationalisation technique and the properties of complex numbers to reach the correct answer. Mistakes can be avoided while finding the products. Remember that $i^2 = -1$.