A) \[\left| \frac{1}{{{z}_{1}}+{{z}_{2}}+{{z}_{3}}+....+{{z}_{n}}} \right|\]
B) \[\left| \frac{1}{{{z}_{1}}}+\frac{1}{{{z}_{2}}}+\frac{1}{{{z}_{3}}}+...+\frac{1}{{{z}_{n}}} \right|\]
C) \[\left| \frac{1}{{{z}_{1}}{{z}_{2}}}+\frac{1}{{{z}_{3}}{{z}_{4}}}+....+\frac{1}{{{z}_{n-1}}{{z}_{n}}} \right|\]
D) \[1\]
Correct Answer: B
Solution :
We have \[|{{z}_{k}}|=1,\,\,\,k=1,2.....n.\] \[\Rightarrow \] \[|{{z}_{k}}{{|}^{2}}=1\,\,\,\,\Rightarrow \,\,\,{{z}_{k}}{{\bar{z}}_{k}}=1\,\,\,\,\,\Rightarrow \,\,\,\,{{\bar{z}}_{k}}=\frac{1}{{{z}_{k}}}\] Therefore, \[|{{z}_{1}}+{{z}_{2}}+....+{{z}_{n}}|=|\overline{{{z}_{1}}+{{z}_{2}}+.....+{{z}_{n}}|}\] \[(\because \,\,|z|\,\,\,=\,\,|\bar{z}|)\] \[=|{{z}_{1}}+{{z}_{2}}+.....+{{\bar{z}}_{n}}|\,\,=\left| \frac{1}{{{z}_{1}}}+\frac{1}{{{z}_{2}}}+....+\frac{1}{{{z}_{n}}} \right|\]
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