A) 0
B) 1
C) \[-1\]
D) None
Correct Answer: A
Solution :
\[{{z}_{1}}{{\bar{z}}_{1}}={{z}_{2}}{{\bar{z}}_{2}}=........={{z}_{n}}{{\bar{z}}_{n}}=1\] \[\Rightarrow \,\,{{\bar{z}}_{1}}=\frac{1}{{{z}_{1}}},{{\bar{z}}_{2}}=\frac{1}{{{z}_{2}}},{{\bar{z}}_{3}}=\frac{1}{{{z}_{3}}},.....,{{\bar{z}}_{n}}=\frac{1}{{{z}_{n}}}\] \[\therefore \,\,\,|{{z}_{1}}+{{z}_{2}}+.......+{{z}_{n}}|-\left| \frac{1}{{{z}_{1}}}+\frac{1}{{{z}_{2}}}+......+\frac{1}{{{z}_{n}}} \right|\] \[\therefore \,\,\,|{{z}_{1}}+{{z}_{2}}+....+{{z}_{n}}|-|{{\bar{z}}_{1}}+{{\bar{z}}_{2}}+....+{{\bar{z}}_{n}}|=0\]
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