Question

Let $z_1=\frac{2\sqrt{3}+i6\sqrt{7}}{6\sqrt{7}+i2\sqrt{3}}$ and $z_2=\frac{\sqrt{11}+i3\sqrt{13}}{3\sqrt{13}+i\sqrt{11}}$
Then, $|\frac{1}{z_1}+\frac{1}{z_2}|$ is equal to :

• A. $47$
• B. $264$
• C. $|z_1-z_2|$
• D. $|z_1+z_2|$
• E. $|z_1{z}_2|$

Answer 1

Answer: (D)

$|z_1+z_2|$

Given, $z_1=\frac{2\sqrt{3}+i6\sqrt{7}}{6\sqrt{7}+i2\sqrt{3}}\times \frac{6\sqrt{7}-i2\sqrt{3}}{6\sqrt{7}-i2\sqrt{3}}$

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

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

and $z_2=\frac{\sqrt{11}+i3\sqrt{13}}{3\sqrt{13}-i\sqrt{11}}\times \frac{3\sqrt{13}+i\sqrt{11}}{3\sqrt{13}+i\sqrt{11}}$

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

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

Now, $\frac{1}{z_1}=\frac{\overline{z_1}}{|z_1{|}^2}=\frac{\frac{\sqrt{21}}{11}-\frac{10i}{11}}{\sqrt{\frac{21}{121}+\frac{100}{121}}}=\frac{\sqrt{21}}{11}-\frac{10i}{11}$ $=\overline{z_1}$

and $\frac{1}{z_2}=\frac{\overline{z_2}}{|\overline{z_2}{|}^2}=\frac{-i}{\sqrt{1}}=-i=\overline{z_2}$

Therefore, $\frac{1}{z_1}+\frac{1}{z_2}=\overline{z_1}+\overline{z_2}=\overline{z_1+z_2}$

$\Rightarrow |\frac{1}{z_1}+\frac{1}{z_2}|=|z_1+z_2|(\therefore |z|=|\overline{z}|)$