• # question_answer If the complex number ${{z}_{1}},{{z}_{2}}$ the origin form an equilateral triangle then  $z_{1}^{2}+z_{2}^{2}=$ [IIT 1983] A) ${{z}_{1}}\,{{z}_{2}}$ B)  ${{z}_{1}}\,\overline{{{z}_{2}}}$ C) $\overline{{{z}_{2}}}\,{{z}_{1}}$ D) $|{{z}_{1}}{{|}^{2}}=|{{z}_{2}}{{|}^{2}}$

Let $OA,OB$ be the sides of an equilateral $\Delta OAB$and let $OA,OB$represent the complex numbers or vectors  ${{z}_{1}},\,{{z}_{2}}$ respectively. From the equilateral $\Delta OAB,\overrightarrow{AB}={{z}_{2}}-{{z}_{1}}$ $\therefore$  $arg\,\,\left( \frac{{{z}_{2}}-{{z}_{1}}}{{{z}_{2}}} \right)=arg\,({{z}_{2}}-{{z}_{1}})-arg\,{{z}_{2}}=\pi /3$ and$arg\left( \frac{{{z}_{2}}}{{{z}_{1}}} \right)=arg({{z}_{2}})-arg({{z}_{1}})=\frac{\pi }{3}$ Also$\left| \frac{{{z}_{2}}-{{z}_{1}}}{{{z}_{2}}} \right|=1=\left| \frac{{{z}_{2}}}{{{z}_{1}}} \right|$, since triangle is equilateral. Thus the vectors $\frac{{{z}_{2}}-{{z}_{1}}}{{{z}_{2}}}$and $\frac{{{z}_{2}}}{{{z}_{1}}}$ have same modulus and same argument, which implies that the vectors are equal, that is  $\frac{{{z}_{2}}-{{z}_{1}}}{{{z}_{2}}}=\frac{{{z}_{2}}}{{{z}_{1}}}$Þ${{z}_{1}}{{z}_{2}}-z_{1}^{2}=z_{2}^{2}$Þ$z_{1}^{2}+z_{2}^{2}={{z}_{1}}{{z}_{2}}$ Note: Students should remember this question as a formula.