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Question

If $$z_1, z_2, z_3$$ are imaginary numbers such that $$|z_1| = |z_2| = |z_3| = \begin{vmatrix}\dfrac{1}{z_1}+\dfrac{1}{z_2}+\dfrac{1}{z_3}\end{vmatrix} = 1$$ then $$|z_1+z_2+z_3|$$ is


A
Equal to 1
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B
Less than 1
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C
Greater than 1
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D
Equal to 3
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Solution

The correct option is A Equal to $$1$$
using $$z\overline z = |z|^2$$
$$z_1\overline{z_1}=|z_1|^2=1\Rightarrow \dfrac{1}{z_1}=\overline{z_1}$$
$$z_2\overline{z_2}=|z_2|^2=1\Rightarrow \dfrac{1}{z_2}=\overline{z_2}$$
and $$z\overline{z_3}=|z_3|^2=1\Rightarrow \dfrac{1}{z_3}=\overline{z_3}$$

Now using, $$\bigg |\dfrac{1}{z_1}+\dfrac{1}{z_2}+\dfrac{1}{z_3}\bigg |=1$$
$$\Rightarrow |\overline z_1+\overline z_2+\overline z_3|=1$$
$$\Rightarrow |\overline{z_1+z_2+z_3}|=1$$
$$\therefore |z_1+z_2+z_3|=1$$, since $$|\overline z|=|z|$$ 

Mathematics

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