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

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

Suggest Corrections

0

Similar questions
View More

People also searched for
View More