A) Equilateral triangle
B) Isosceles triangle
C) Right angled triangle
D) None of these
Correct Answer: A
Solution :
The cube roots of unity are\[1,\omega ,{{\omega }^{2}}\]. We know that if \[{{z}_{1}},{{z}_{2}},{{z}_{3}}\] are the vertices of an equilateral triangle in the Argand plane. Then \[z_{1}^{2}+z_{2}^{2}+z_{3}^{2}={{z}_{1}}{{z}_{2}}+{{z}_{2}}{{z}_{3}}+{{z}_{3}}{{z}_{1}}\] If we take \[{{z}_{1}}=1,{{z}_{2}}=\omega ,{{z}_{3}}={{\omega }^{2}}\] Then \[z_{1}^{2}+z_{2}^{2}+z_{3}^{2}=1+{{\omega }^{2}}+{{\omega }^{4}}=0\] and \[{{z}_{1}}{{z}_{2}}+{{z}_{2}}{{z}_{3}}+{{z}_{3}}{{z}_{1}}=1.\omega +\omega .{{\omega }^{2}}+{{\omega }^{2}}.1\] \[=\omega +{{\omega }^{3}}+{{\omega }^{2}}=1+\omega +{{\omega }^{2}}=0\] Thus \[z_{1}^{2}+z_{2}^{2}+z_{3}^{2}={{z}_{1}}{{z}_{2}}+{{z}_{2}}{{z}_{3}}+{{z}_{3}}{{z}_{1}}\] Therefore triangle is equilateral. Note: Students should remember this question as a fact.
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