A) 1
B) \[|{{z}_{1}}|+|{{z}_{2}}|+.......+|{{z}_{n}}|\]
C) \[\left| \frac{1}{{{z}_{1}}}+\frac{1}{{{z}_{2}}}+.........+\frac{1}{{{z}_{n}}} \right|\]
D) None of these
Correct Answer: C
Solution :
We have \[|{{z}_{k}}|=1,k=1,\,\,2,....n\] Þ \[|{{z}_{k}}{{|}^{2}}=1\,\,\Rightarrow {{z}_{k}}{{\overline{z}}_{k}}=1\,\Rightarrow {{\overline{z}}_{k}}=\frac{1}{{{z}_{k}}}\] Therefore \[|{{z}_{1}}+{{z}_{2}}+....+{{z}_{n}}|=|\overline{{{z}_{1}}+{{z}_{2}}+....+{{z}_{n}}}|\] \[(\because \,\,\,|z|\,=\,|\,\overline{z}|)\] \[=|{{\overline{z}}_{1}}+\overline{{{z}_{2}}}+.....+{{\overline{z}}_{n}}|=\left| \frac{1}{{{z}_{1}}}+\frac{1}{{{z}_{2}}}+....+\frac{1}{{{z}_{n}}} \right|\] Aliter: Let \[{{z}_{k}}=\cos {{\theta }_{k}}+i\sin {{\theta }_{k}},\,\,\,k=1,\,\,2,....n\] So that \[|{{z}_{k}}|=\sqrt{{{\cos }^{2}}{{\theta }_{k}}+{{\sin }^{2}}{{\theta }_{k}}}=1\] Then \[\frac{1}{{{z}_{k}}}={{(\cos {{\theta }_{k}}+i\sin {{\theta }_{k}})}^{-1}}=(\cos {{\theta }_{k}}-i\sin {{\theta }_{k}})\] Now, \[{{z}_{1}}+{{z}_{2}}+.....+{{z}_{n}}\] \[=(\cos {{\theta }_{1}}+.....+\cos {{\theta }_{n}})-i(\sin {{\theta }_{1}}+.....+\sin {{\theta }_{n}})\] and \[\left( \frac{1}{{{z}_{1}}} \right)+\left( \frac{1}{{{z}_{2}}} \right)+.....+\left( \frac{1}{{{z}_{n}}} \right)\] \[=(\cos {{\theta }_{1}}+.....+\cos {{\theta }_{n}})-i(\sin {{\theta }_{1}}+.....+\sin {{\theta }_{n}})\] Hence \[|{{z}_{1}}+{{z}_{2}}+.....+{{z}_{n}}|=\left| \frac{1}{{{z}_{1}}}+\frac{1}{{{z}_{2}}}+.....+\frac{1}{{{z}_{n}}} \right|\] Since each side is equal to \[\sqrt{{{(\cos {{\theta }_{1}}+.....+\cos {{\theta }_{n}})}^{2}}+{{(\sin {{\theta }_{1}}+....+\sin {{\theta }_{n}})}^{2}}}\]
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