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) | |
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