**If \(z_1 , z_2 , z_3 , z_4 \in \mathbb{C} \) satisfy \(z_1 + z_2 + z_3 + z_4 = 0 \) and \(|z_1|^2 + |z_2|^2 + |z_3|^2 + |z_4|^2 = 1 \) then the least value of \(|z_1 – z_2 |^2 + |z_1 – z_4|^2 + |z_2 – z_3|^2 + |z_3 – z_4|^2 \) is 2**

**True**

Discussion:

\(|z_1 – z_2|^2 = (z_1 – z_2)(\bar{z_1} – \bar{z_2}) = |z_1|^2 + |z_2|^2 – (z_1 \bar {z_2} + \bar {z_1} {z_2} ) \)

Similarly we compute the others to get the total sum as

\(2 ( |z_1|^2 + |z_2|^2 + |z_3|^2 + |z_4|^2 ) \) – \((z_1 + z_3) (\bar{z_2} + \bar {z_4} ) \)

– \(( \bar {z_1} + \bar {z_3} ) (z_2 + z_4 ) \)

Since \(z_1 + z_3 = – (z_2 + z_4) \) thus \(\bar {z_2 } + \bar {z_4} = – ( \bar {z_1} + \bar {z_3} ) \) the above expression reduces to

\(2 ( |z_1|^2 + |z_2|^2 + |z_3|^2 + |z_4|^2 ) + 2 |z_1 + z_3|^2 \ge 2 \)