If z 1- z 2- z 3 are imaginary numbers such that - z 1- z 2- z 3-1- z 1-1- z 2-1- z 3-1 then - z 1- z 2- z 3-A- equal to 1B- less than 1C- greater than 1D- equal to 3

|z|=1⇒ zz̅=1⇒z̅=1 z |z_1+z_2+z_3|=|(z_1+z_2+z_3)| =|1z_1+1z_2+1z_3|=1 ...

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Equality of 2 Complex Numbers

If z 1, z 2, ...

Question

If $z_{1}, z_{2}, z_{3}$ are imaginary numbers such that $| z_{1} | = | z_{2} | = | z_{3} | = ∣ ∣ ∣ \frac{1}{z_{1}} + \frac{1}{z_{2}} + \frac{1}{z_{3}} ∣ ∣ ∣ = 1$ then $| z_{1} + z_{2} + z_{3} |$ is

equal to

1

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less than

1

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greater than

1

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equal to

3

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z_{1}, z_{2}, z_{3}

| z_{1} | = | z_{2} | = | z_{3} | = ∣ ∣ \frac{1}{z_{1}} + \frac{1}{z_{2}} + \frac{1}{z_{3}} ∣ ∣ = 1

| z_{1} + z_{2} + z_{3} |

if $z_{1}, z_{2} a n d z_{3}$ are complex numbers such that

| $z_{1}$ |=| $z_{2}$ |=| $z_{3}$ |= $∣ ∣ \frac{1}{z_{1}} + \frac{1}{z_{2}} + \frac{1}{z_{3}} ∣ ∣ = 1$ ,

then | $z_{1}$ + $z_{2}$ + $z_{3}$ |

If $z_{1}, z_{2} a n d z_{3}$ are complex numbers such that

| $z_{1}$ |=| $z_{2}$ |=| $z_{3}$ |= $∣ ∣ \frac{1}{z_{1}} + \frac{1}{z_{2}} + \frac{1}{z_{3}} ∣ ∣ = 1$ ,

then | $z_{1}$ + $z_{2}$ + $z_{3}$ |

z_{1}, z_{2}, z_{3}

| z_{1} | = | z_{2} | = | z_{3} | = ∣ ∣ ∣ \begin{matrix} \frac{1}{z_{1}} + \frac{1}{z_{2}} + \frac{1}{z_{3}} \end{matrix} ∣ ∣ ∣ = 1

| z_{1} + z_{2} + z_{3} |

If $Z_{1}, Z_{2}, Z_{3}$ are complex numbers such that

$| Z_{1} | = | Z_{2} | = | Z_{3} | = | \frac{1}{Z_{1}} + \frac{1}{Z_{2}} + \frac{1}{Z_{3}} | = 1$ then $| Z_{1} + | Z_{2} + Z_{3} |$ is

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