TIFR-2012-Maths-D: 39

Question

True/False Question:

If   $z_{1},z_{2},z_{3},z_{4}\in \mathbb{C}$ satisfy $z_{1}+z_{2}+z_{3}+z_{4}=0$ and  $\left | z_{1} \right |^{2}+\left | z_{2} \right |^{2}+\left | z_{3} \right |^{2}+\left | z_{4} \right |^{2}=1$, then the least value of   $\left | z_{1} -z_{2}\right |^{2}+\left | z_{1} -z_{4}\right |^{2}+\left | z_{2}-z_{3} \right |^{2}+\left | z_{3} -z_{4}\right |^{2}$ is $2$.

Answer 1

sqr(|z1 – z2|)+sqr(|z1 – z4|)+sqr(|z2 – z3|)+sqr(| z3 – z4|) >=2

2(sqr(|z1|)+sqr(|z2|)+sqr(|z3|)+sqr(|z4|) ) – 2(|z1| . |z2| +  |z2| .|z3| + |z3|.|z4| + |z4| . |z1|)    .:  sqr(a-b) = sqr(a) + sqr(b) – 2ab .

Given  |z1|2+|z2|2+|z3|2+|z4|2=1

2(1) – 2(|z1| . |z2| +  |z2| .|z3| + |z3|.|z4| + |z4| . |z1|)

Given z1+z2+z3+z4 =0

sqr on both sides  (z1)2 +(z2)2+(z3)2+(z4)2 = – 2( z1 . z2 + z3 . z4)

as we know  |z1|2+|z2|2+|z3|2+|z4|2=1  then  – 1 / 2=  (z1 .z2 +z3 .z4)   silmlarly z4.z1 +z2.z3 = – 1/2

2(1) – 2( – (1/2)+( – (1/2) )  =>  2 + 2