question_answer

If complex numbers \[{{z}_{1}},{{z}_{2}}\,\text{and }{{z}_{3}}\] represent the vertices A, B and C respectively of an isosceles triangle ABC of which \[\angle C\] is right angle, then correct statement is [RPET 1999]

A) \[{{z}_{1}}^{2}+{{z}_{2}}^{2}+{{z}_{3}}^{2}={{z}_{1}}{{z}_{2}}{{z}_{3}}\]

B) \[{{({{z}_{3}}-{{z}_{1}})}^{2}}={{z}_{3}}-{{z}_{2}}\]

C) \[{{({{z}_{1}}-{{z}_{2}})}^{2}}=({{z}_{1}}-{{z}_{3}})\,({{z}_{3}}-{{z}_{2}})\]

D) \[{{({{z}_{1}}-{{z}_{2}})}^{2}}=2({{z}_{1}}-{{z}_{3}})\,({{z}_{3}}-{{z}_{2}})\]

Correct Answer: D

Solution :

\[BC=AC\] and \[\angle C=\pi /2\] By rotation about \[C\] in anti-clockwise sense \[CB=CA\,{{e}^{i\,\pi /2}}\] Þ \[({{z}_{2}}-{{z}_{3}})=({{z}_{1}}-{{z}_{3}})\,{{e}^{i\,\pi /2}}=i\,({{z}_{1}}-{{z}_{3}})\] Þ  \[{{({{z}_{2}}-{{z}_{3}})}^{2}}=-{{({{z}_{1}}-{{z}_{3}})}^{2}}\] Þ \[z_{2}^{2}+z_{3}^{2}-2{{z}_{2}}{{z}_{3}}=-z_{1}^{2}-z_{3}^{2}+2{{z}_{1}}{{z}_{3}}\] Þ \[z_{1}^{2}+z_{2}^{2}-2{{z}_{1}}{{z}_{2}}=2{{z}_{1}}{{z}_{3}}+2{{z}_{2}}{{z}_{3}}-2z_{3}^{2}-2{{z}_{1}}{{z}_{2}}\] Þ \[{{({{z}_{1}}-{{z}_{2}})}^{2}}=2\,[({{z}_{1}}{{z}_{3}}-z_{3}^{2})-({{z}_{1}}{{z}_{2}}-{{z}_{2}}{{z}_{3}})]\] Þ \[{{({{z}_{1}}-{{z}_{2}})}^{2}}=2({{z}_{1}}-{{z}_{3}})\,({{z}_{3}}-{{z}_{2}})\].