Question

Consider the following two statements:
Statement I: For any two non-zero complex numbers \( z_1, z_2 \),
\((|z_1| + |z_2|) \left| \frac{z_1}{|z_1|} + \frac{z_2}{|z_2|} \right| \leq 2 (|z_1| + |z_2|)\)
Statement II: If \( x, y, z \) are three distinct complex numbers and \( a, b, c \) are three positive real numbers such that  
\(\frac{a}{|y - z|} = \frac{b}{|z - x|} = \frac{c}{|x - y|},\)
then  
\(\frac{a^2}{y - z} + \frac{b^2}{z - x} + \frac{c^2}{x - y} = 1.\)
Between the above two statements,

• A. both Statement I and Statement II are incorrect.
• B. Statement I is incorrect but Statement II is correct.
• C. Statement I is correct but Statement II is incorrect.
• D. both Statement I and Statement II are correct.

Answer 1

The Correct Option is C

Statement I:

\[ \left( \lvert z_1 \rvert + \lvert z_2 \rvert \right) \left\lvert \frac{z_1}{\lvert z_1 \rvert} + \frac{z_2}{\lvert z_2 \rvert} \right\rvert \leq 2 \left( \lvert z_1 \rvert + \lvert z_2 \rvert \right) \]

Since

\[ \left\lvert \frac{z_1}{\lvert z_1 \rvert} + \frac{z_2}{\lvert z_2 \rvert} \right\rvert \leq 2 \]

we have

\[ \left( \lvert z_1 \rvert + \lvert z_2 \rvert \right) \left\lvert \frac{z_1}{\lvert z_1 \rvert} + \frac{z_2}{\lvert z_2 \rvert} \right\rvert \leq 2 \left( \lvert z_1 \rvert + \lvert z_2 \rvert \right) \]

Thus, Statement I is correct.

Statement II: Given

\[ \frac{a}{\lvert y - z \rvert} = \frac{b}{\lvert z - x \rvert} = \frac{c}{\lvert x - y \rvert} \]

let

\[ \frac{a}{\lvert y - z \rvert} = \frac{b}{\lvert z - x \rvert} = \frac{c}{\lvert x - y \rvert} = \lambda \]

Then,

\[ a^2 = \lambda \lvert y - z \rvert, \quad b^2 = \lambda \lvert z - x \rvert, \quad c^2 = \lambda \lvert x - y \rvert \]

Substituting, we get:

\[ \frac{a^2}{y - z} + \frac{b^2}{z - x} + \frac{c^2}{x - y} = \lambda \left( \frac{y - z}{y - z} + \frac{z - x}{z - x} + \frac{x - y}{x - y} \right) \]
Thus, Statement II is false.

Answer 2

Step 1: Analyze Statement I
For non-zero complex numbers z₁, z₂, apply the triangle inequality to the unit-direction terms: \[ \left|\frac{z_1}{|z_1|} + \frac{z_2}{|z_2|}\right| \le \left|\frac{z_1}{|z_1|}\right| + \left|\frac{z_2}{|z_2|}\right| = 1 + 1 = 2. \] Multiplying both sides by \((|z_1| + |z_2|)\) gives \[ (|z_1| + |z_2|)\left|\frac{z_1}{|z_1|} + \frac{z_2}{|z_2|}\right| \le 2(|z_1| + |z_2|), \] which is exactly Statement I. Hence, Statement I is true. Equality holds when \(\frac{z_1}{|z_1|}\) and \(\frac{z_2}{|z_2|}\) have the same argument (i.e., z₁ and z₂ point in the same direction).

Step 2: Analyze Statement II
Given distinct complex numbers x, y, z and positive reals a, b, c such that \[ \frac{a}{|y-z|} = \frac{b}{|z-x|} = \frac{c}{|x-y|} = k \; (>0), \] we have \[ a = k|y-z|,\quad b = k|z-x|,\quad c = k|x-y|. \] Then \[ \frac{a^2}{y-z} + \frac{b^2}{z-x} + \frac{c^2}{x-y} = k^2\left(\frac{|y-z|^2}{y-z} + \frac{|z-x|^2}{z-x} + \frac{|x-y|^2}{x-y}\right). \] Using \(|w|^2 = w\overline{w}\) and \(w \ne 0\) (since the points are distinct), we get \(\frac{|w|^2}{w} = \overline{w}\). Hence \[ \frac{a^2}{y-z} + \frac{b^2}{z-x} + \frac{c^2}{x-y} = k^2\big(\overline{y - z} + \overline{z - x} + \overline{x - y}\big) = k^2\cdot \overline{(y - z + z - x + x - y)} = k^2\cdot \overline{0} = 0. \] Therefore, the expression equals 0 (for all such choices), not 1. Hence, Statement II is false.

Conclusion

Statement I is correct but Statement II is incorrect.