Consider the following two statements :
Statement I: For any two non-zero complex numbers z₁, z₂,(|z₁|+|z₂|)|z₁/|z₁|+z₂/|z₂|| ≤ 2(|z₁|+|z₂|) , and
Statement II : If x, y, z are three distinct complex numbers and a, b, c are three positive real numbers such that a/(|y-z|)=b/(|z-x|)=c/(|x-y|), then a²/(y-z)+b²/(z-x)+c²/(x-y)=1.
Between the above two statements,

Solution

Maths sir · 3 steps

Divide each number by its modulus to get unit vectors, and use |w|² = ww̄ to flatten Statement II.

1. Statement I. z₁/|z₁| and z₂/|z₂| both have modulus 1. By the triangle inequality, their sum has modulus at most 1 + 1 = 2. Multiplying both sides by the positive number |z₁|+|z₂| gives the stated inequality, so Statement I is correct.
2. Rewrite a squared. Call the common ratio k, so a = k|y-z|. Then a² = k²|y-z|² = k²(y-z)(y-z)̅, which gives a²/(y-z) = k² (y-z)̅. The same works for b and c.

   a²/(y-z)+b²/(z-x)+c²/(x-y) = k² (y-z)+(z-x)+(x-y)̅ = k² × 0̄ = 0

3. So. The sum in Statement II is 0, not 1, so Statement II is incorrect. Statement I is correct and Statement II is incorrect, which is option (b).

The idea to keep. Whenever you see |w|² in a numerator over w, write |w|² = ww̄ and cancel. Here the three differences add to zero, so the whole sum is zero.

Answer from the NTA final answer key. Solution by PrepWiser.