question_answer

Let \[{{z}_{1}},{{z}_{2}},{{z}_{3}},{{z}_{4}}\] be the vertices A, B, C, D respectively of a square on the argand plane taken in anticlockwise direction, then -

A) \[2{{z}_{4}}=(1-i)\,\,{{z}_{1}}+(1+i)\,\,{{z}_{1}}\]

B) \[2{{z}_{1}}=(1-i)\,\,{{z}_{1}}+(1+i)\,\,{{z}_{3}}\]

C) \[2{{z}_{4}}=(1+i)\,\,{{z}_{1}}+(1-i)\,\,{{z}_{3}}\]

D) \[3{{z}_{1}}=(1+i)\,\,{{z}_{1}}+(1-i)\,\,{{z}_{3}}\]

Correct Answer: A

Solution :

[A]

\[\frac{{{z}_{4}}-{{z}_{1}}}{{{z}_{2}}-{{z}_{1}}}=i\]
\[i.e.\,\,\,\,{{z}_{4}}-{{z}_{1}}=i\,({{z}_{2}}-{{z}_{1}})\]
Also \[{{z}_{1}}+{{z}_{3}}={{z}_{4}}+{{z}_{2}}\]
i.e.        \[{{z}_{1}}+{{z}_{3}}={{z}_{1}}+i\,({{z}_{2}}-{{z}_{1}})+{{z}_{2}}\]
i.e.        \[{{z}_{3}}+i{{z}_{1}}=(1+i)\,{{z}_{2}}\]
i.e.        \[2{{z}_{2}}=(1-i)\,\,{{z}_{3}}+(1-i)\,\,i\,\,{{z}_{1}}\]
\[=(1+i)\,\,{{z}_{1}}+(1-i)\,\,{{z}_{3}}\]		?.(i)
\[2\,\,({{z}_{1}}+{{z}_{3}}-{{z}_{4}})=(1+i)\,\,{{z}_{1}}+(1-i)\,\,{{z}_{3}}\]
\[\therefore \,\,\,\,\,\,\,\,\,\,\,2{{z}_{4}}=2{{z}_{1}}+2{{z}_{3}}-(1+i)\,\,{{z}_{1}}-(1-i)\,\,{{z}_{3}}\]
\[=(1-i)\,\,{{z}_{1}}+(1+i)\,\,{{z}_{3}}\]	?(ii)