question_answer

Let the complex numbers \[{{z}_{1}},{{z}_{2}}\] and \[{{z}_{3}}\] be the vertices of an equilateral triangle. Let \[{{z}_{0}}\]be the circumcentre of the triangle, then \[z_{1}^{2}+z_{2}^{2}+z_{3}^{2}=\] [IIT 1981]

A) \[z_{0}^{2}\]

B) \[-z_{0}^{2}\]

C) \[3z_{0}^{2}\]

D) \[-3z_{0}^{2}\]

Correct Answer: C

Solution :

Let \[r\] be the circum radius of the equilateral triangle and \[\omega \] the cube root of unity. Let \[ABC\] be the equilateral triangle with \[{{z}_{1}},{{z}_{2}}\] and \[{{z}_{3}}\] as its vertices \[A,B\]and C respectively with circumcentre \[{O}'({{z}_{0}})\]. The vectors \[{O}'A,{O}'B,{O}'C\]are equal and parallel to \[O{A}',O{B}',O{C}'\] respectively. Then the vectors \[\overrightarrow{O{A}'}={{z}_{1}}-{{z}_{0}}=r{{e}^{i\theta }}\] \[\overrightarrow{O{B}'}={{z}_{2}}-{{z}_{0}}=r{{e}^{\left( \theta +\frac{2\pi }{3} \right)}}=r\omega {{e}^{i\theta }}\] \[\overrightarrow{O{C}'}={{z}_{3}}-{{z}_{0}}=r{{e}^{i\,\left( \theta +\frac{4\pi }{3} \right)}}=r{{\omega }^{2}}{{e}^{i\theta }}\] \[\therefore \] \[{{z}_{1}}={{z}_{0}}+r{{e}^{i\theta }},{{z}_{2}}={{z}_{0}}+r\omega {{e}^{i\theta }},{{z}_{3}}={{z}_{0}}+r{{\omega }^{2}}{{e}^{i\theta }}\] Squaring and adding \[z_{1}^{2}+z_{2}^{2}+z_{3}^{2}=3z_{0}^{2}+2(1+\omega +{{\omega }^{2}}){{z}_{0}}r{{e}^{i\theta }}\]+ \[(1+{{\omega }^{2}}+{{\omega }^{4}}){{r}^{2}}{{e}^{i2\theta }}\] \[=3z_{^{0}}^{2},\]since \[1+\omega +{{\omega }^{2}}=0=1+{{\omega }^{2}}+{{\omega }^{4}}\] Note: Students should remember this question as a formula.