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) |
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