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

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

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

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

Correct Answer: A

Solution :

\[\frac{{{z}_{4}}-{{z}_{1}}}{{{z}_{2}}-{{z}_{1}}}=i\] 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)