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If $z_{1}, z_{2}, z_{3}$ are imaginary numbers such that $\left|z_{1}\right|=\left|z_{2}\right|=\left|z_{3}\right|=\left|\frac{1}{z_{1}}+\frac{1}{z_{2}}+\frac{1}{z_{3}}\right|=1,$ then
$\left|z_{1}+z_{2}+z_{3}\right|$ is
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$\left|z_{1}+z_{2}+z_{3}\right|$ is
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Verified Answer
The correct answer is:
equal to 1
We have, $\left|z_{1}\right|=\left|z_{2}\right|=\left|z_{3}\right|=1$
$\Rightarrow \quad\left|z_{1}\right|^{2}=\left|z_{2}\right|^{2}=\left|z_{3}\right|^{2}=1$
$\Rightarrow \quad \quad z_{1} \bar{z}_{1}=z_{2} \bar{z}_{2}=z_{3} \bar{z}_{3}=1$
$\Rightarrow \quad \frac{1}{z_{1}}=\bar{z}_{1} \frac{1}{z_{2}}=\bar{z}_{2} \frac{1}{z_{3}}=\bar{z}_{3}$
Now. $\quad\left|\frac{1}{z_{1}}+\frac{1}{z_{2}}+\frac{1}{z_{3}}\right|=1$
$\begin{array}{ll}\Rightarrow & \left|\bar{z}_{1}+\bar{z}_{2}+\bar{z}_{3}\right|=1 \\ \Rightarrow & \left|\overline{z_{1}+z_{2}+z_{3}}\right|=1 \\ \therefore & \left|z_{1}+z_{2}+z_{j}\right|=1\end{array}$
$\Rightarrow \quad\left|z_{1}\right|^{2}=\left|z_{2}\right|^{2}=\left|z_{3}\right|^{2}=1$
$\Rightarrow \quad \quad z_{1} \bar{z}_{1}=z_{2} \bar{z}_{2}=z_{3} \bar{z}_{3}=1$
$\Rightarrow \quad \frac{1}{z_{1}}=\bar{z}_{1} \frac{1}{z_{2}}=\bar{z}_{2} \frac{1}{z_{3}}=\bar{z}_{3}$
Now. $\quad\left|\frac{1}{z_{1}}+\frac{1}{z_{2}}+\frac{1}{z_{3}}\right|=1$
$\begin{array}{ll}\Rightarrow & \left|\bar{z}_{1}+\bar{z}_{2}+\bar{z}_{3}\right|=1 \\ \Rightarrow & \left|\overline{z_{1}+z_{2}+z_{3}}\right|=1 \\ \therefore & \left|z_{1}+z_{2}+z_{j}\right|=1\end{array}$
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