Question
Let θ 1 , θ 2 , … , θ 10 be positive valued angles (in radian) such that θ 1 + θ 2 + ⋯ + θ 10 = 2 π . Define the complex numbers z 1 = e i θ 1 , z k = z k - 1 e i θ k for k = 2 , 3 , … , 10 , where i = - 1 . Consider the statements P and Q given below: P : z 2 - z 1 + z 3 - z 2 + ⋯ + z 10 - z 9 + z 1 - z 10 ≤ 2 π Q : z 2 2 - z 1 2 + z 3 2 - z 2 2 + ⋯ + z 10 2 - z 9 2 + z 1 2 - z 10 2 ≤ 4 π
Step-by-step solution
Given, θ 1 + θ 2 + . . . . . + θ 10 = 2 π , z 1 = e i θ 1 ,   z k = z k - 1 e i θ k Now, z k = z k - 1 e i θ k ⇒ z 2 = z 1 e i θ 1 ,   z 3 = z 2 e i θ 2 ,   … … As, z 1 = e i θ 1 ⇒ z 1 = 1 z 2 = z 1 e i θ ⇒ z 2 = z 1 = 1 Similarly z 1 = z 2 … = z k = 1 ⇒ z 1 , z 2 , … , z k lies on a circle of unit radius. We know, z k - z k - 1 represents a line segment joining z k   &   z k - 1 . Both z k   &   z k - 1 lies on a unit circle Since θ 1 + θ 2 + . . . . + θ 10 = 2 π We get So, z 2 - z 1 , z 3 - z 2 , … , z 1 - z 10 are the sides of a decagon circumscribed by a circle of unit radius We know, the sum of the length of sides of a decagon is less than the circumference of its circumcircle. ⇒ z 2 - z 1 + z 3 - z 2 + ⋯ + z 10 - z 9 + z 1 - z 10 ≤ 2 π Hence, P is true. Similarly, z 1 2 = e i 2 θ 1 ,   z k 2 = z k - 1 2 e i 2 θ k ,   … … So, z 2 2 - z 1 2 ,   z 3 2 - z 2 2 ,   z 4 2 - z 3 2 ,   z 5 2 - z 4 2 ,   z 6 2 - z 5 2 are the sides of a pentagon circumscribed by a circle of unit radius So, z 2 2 - z 1 2 + z 3 2 - z 2 2 + … z 6 2 - z 5 2 ≤ 2 π Similarly, z 7 2 - z 8 2 + … + z 1 2 - z 10 2 ≤ 2 π Adding these equations, we get z 2 2 - z 1 2 + z 3 2 - z 2 2 + ⋯ + z 10 2 - z 9 2 + z 1 2 - z 10 2 ≤ 4 π Hence, Q is true.