Question

For $Z_1=\sqrt[6]{6\frac{1-i}{1+i\sqrt{3}}};Z_2=\sqrt[6]{6\frac{1-i}{\sqrt{3}+i}};Z_3=\sqrt[6]{6\frac{1+i}{\sqrt{3}-i}}$ which of the following holds good?

• A. $\sum |Z_1{|}^2=\frac{3}{2}$
• B. $|Z_1{|}^4+|Z_2{|}^4=|Z_3{|}^{-8}$
• C. $\sum |Z_1{|}^3+|Z_2{|}^3=|Z_3{|}^{-6}$
• D. $|Z_1{|}^4+|Z_2{|}^4=|Z_3{|}^8$

Answer 1

The correct option is B $|Z_1{|}^4+|Z_2{|}^4=|Z_3{|}^{-8}$

$z_1=\sqrt{\frac{1-i}{1+i\sqrt{3}}},z_2=\sqrt{\frac{1-i}{\sqrt{3}+i}},z_3=\sqrt{\frac{1+i}{\sqrt{3}-1}}$

$z_1=\sqrt[6]{6\frac{1-i}{1+\sqrt{3}}}=\sqrt[6]{6\frac{(1-i)(1-i\sqrt{3})}{1+3}}=\sqrt[6]{6\frac{1(1-\sqrt{3})-i(1+\sqrt{3})}{4}}$

$|z_1{|}^2=z_1{\overline{z}}_1=\sqrt[6]{6\frac{(1-\sqrt{3})}{4}}\times \sqrt[6]{6\frac{(1-\sqrt{3})+i(1+\sqrt{3})}{4}}$

$|z_1{|}^2=\sqrt[6]{6\frac{(1-\sqrt{3}{)}^2+(1+\sqrt{3}{)}^2}{16}}$

$|z_1{|}^2=\sqrt[6]{6\frac{8}{16}}=\frac{1}{\left(2\right)1/6}$

$z_2=\sqrt[6]{6\frac{1-i}{\sqrt{3}+i}}=\sqrt[6]{6\frac{(1-i)(\sqrt{3}-i)}{(3+1)}}=\sqrt[6]{6\frac{(\sqrt{3}-1)-i(1+\sqrt{3})}{4}}$

$|z_2{|}^2=z_2{\overline{z}}_2=\sqrt[6]{6\frac{(\sqrt{3}-1)-i(1+\sqrt{3})+i(1+\sqrt{3})}{4}}$

$|z_2{|}^2=\sqrt[6]{6\frac{(\sqrt{3}-1{)}^2+(1+\sqrt{3}{)}^2}{16}}=\sqrt{\frac{8}{16}}=\frac{1}{\left(2\right)1/6}$

$z_3=z_3{\overline{z}}_3=\sqrt[6]{6\frac{(\sqrt{3}-1)+(1+\sqrt{3})}{4}\times \frac{(\sqrt{3}-1)-i(1+\sqrt{3})}{4}}$

$=\sqrt[6]{6\frac{(\sqrt{3}-1{)}^2+(1+\sqrt{3}{)}^2}{16}}=\sqrt{\frac{8}{16}}=\frac{1}{\left(2\right)1/6}$

$|z_1{|}^4=\frac{1}{2^{2/6}}|z_2{|}^4=\frac{1}{2^{2/6}}$

$|z_3{|}^8=\frac{1}{2^{4/6}}\Rightarrow |z_3{|}^{-8}=2^{4/6}$

$\Rightarrow |z_1{|}^4+|z_2{|}^4=\frac{1}{2^{2/6}}+\frac{1}{2^{2/6}}=\frac{2}{2^{2/6}}=2^{4/6}$

$=|z_3{|}^{-8}$

so, $|z_1{|}^4+|z_2{|}^4=|z_3{|}^{-8}$ as

so, option $\left(B\right)$ is right.