Question

Let ${\text{z}}_1={\displaystyle \frac{2\sqrt{3}+i6\sqrt{7}}{6\sqrt{7}+i2\sqrt{3}}}$ and ${\text{z}}_2={\displaystyle \frac{\sqrt{11}+i3\sqrt{13}}{3\sqrt{13}+i\sqrt{11}}}$, then, $\left|{\displaystyle \frac{1}{z_1}}+{\displaystyle \frac{1}{z_2}}\right|$ is equal to

$$(a)47$$

$(b)264$

$(c)\left|z_1-z_2\right|$

$(d)\left|z_1+z_2\right|$

$(e)\left|z_1{z}_2\right|$

Answer 1

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

Complete step-by-step answer:

Rationalising $z_1$,

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

We get,

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

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

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

Now,

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

Now, rationalising $z_2$,

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

We get,

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

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

$={\displaystyle \frac{128i}{128}}=i$

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

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

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

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

$\therefore {\displaystyle \frac{1}{z_1}}+{\displaystyle \frac{1}{z_2}}=\overline{{\text{z}}_1}+\overline{{\text{z}}_2}=\overline{z_1+z_2}$

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

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 $i^2=-1$.