Q.

Let θ1,θ2,…,θ10 be positive valued angles (in radian) such that θ1+θ2+⋯+θ10=2π. Define the complex numbers z1=eiθ1, zk=zk-1eiθk for k=2,3,…,10,
where i=-1. Consider the statements P and Q given below:

P|z2-z1|+|z3-z2|+⋯+|z10-z9|+|z1-z10|≤2π

Q|z22-z12|+|z32-z22|+⋯+|z102-z92|+|z12-z102|≤4π                        [2021]

1 P is TRUE and Q is FALSE    
2 Q is TRUE and P is FALSE    
3 both P and Q are TRUE    
4 both P and Q are FALSE  

Ans.

(3)

Since, |z1|=|z2|=⋯=|z10|=1

θ2=arc(z1,z2)

|z2-z1|=length of line AB≤length of arc AB

|z3-z2|=length of line BC≤length of arc BC

∴ Sum of length of these 10 lines ≤ Sum of length of arcs (i.e. 2π)

                                                                   [∵ θ1+θ2+θ3+⋯+θ10=2π]

∴ P: |z2-z1|+|z3-z2|+⋯+|z1-z10|≤2π

P is true.

Now, |z22-z12|=|z2-z1||z2+z1|

We know that 

|z2+z1|≤|z2|+|z1|≤2

∴ |z22-z12|+|z32-z22|+⋯+|z12-z102|≤2{|z2-z1|+|z3-z2|+⋯+|z1-z10|}≤2(2π)⇒Q≤4π

Q is also true.