Key Concepts and Formulas

• Triangle Inequality: For complex numbers $z_1$ and $z_2$, $|z_1 + z_2| \leq |z_1| + |z_2|$.
• Modulus and Conjugate: For a complex number $z$, $|z|^2 = z\bar{z}$, where $\bar{z}$ is the complex conjugate of $z$. Also, $\overline{z_1 - z_2} = \bar{z_1} - \bar{z_2}$.
• Polar Form: A complex number $z$ can be written as $z = |z|e^{i\theta}$, where $|z|$ is the modulus and $\theta$ is the argument.

Step-by-Step Solution

1. Analysis of Statement I

Step 1: Simplify the given inequality. The given inequality is $(|z_1|+|z_2|)\left|\frac{z_1}{\left|z_1\right|}+\frac{z_2}{\left|z_2\right|}\right| \leq 2\left(\left|z_1\right|+\left|z_2\right|\right)$. Since $|z_1|+|z_2|$ is a positive real number, we can divide both sides by it without changing the inequality's direction: $\left|\frac{z_1}{\left|z_1\right|}+\frac{z_2}{\left|z_2\right|}\right| \leq 2$ Explanation: This simplification isolates the core of the inequality for easier analysis.

Step 2: Recognize unit complex numbers. Let $u_1 = \frac{z_1}{|z_1|}$ and $u_2 = \frac{z_2}{|z_2|}$. Then $|u_1| = |u_2| = 1$, so $u_1$ and $u_2$ are unit complex numbers. Explanation: Identifying these as unit complex numbers is key because we know their moduli are equal to 1.

Step 3: Apply the Triangle Inequality. We need to check if $|u_1 + u_2| \leq 2$ is always true. By the triangle inequality, $|u_1 + u_2| \leq |u_1| + |u_2| = 1 + 1 = 2$. Thus, $|u_1 + u_2| \leq 2$ is always true. Explanation: The triangle inequality directly shows that the modulus of the sum of two unit complex numbers is always less than or equal to 2.

Step 4: Alternative Approach using Polar Form. Let $z_1 = |z_1|e^{i\theta_1}$ and $z_2 = |z_2|e^{i\theta_2}$. Then $\frac{z_1}{|z_1|} = e^{i\theta_1}$ and $\frac{z_2}{|z_2|} = e^{i\theta_2}$. We want to evaluate $|e^{i\theta_1} + e^{i\theta_2}|$. $|e^{i\theta_1} + e^{i\theta_2}|^2 = (e^{i\theta_1} + e^{i\theta_2})(\overline{e^{i\theta_1} + e^{i\theta_2}}) = (e^{i\theta_1} + e^{i\theta_2})(e^{-i\theta_1} + e^{-i\theta_2})$ $= e^{i\theta_1}e^{-i\theta_1} + e^{i\theta_1}e^{-i\theta_2} + e^{i\theta_2}e^{-i\theta_1} + e^{i\theta_2}e^{-i\theta_2} = 1 + e^{i(\theta_1 - \theta_2)} + e^{i(\theta_2 - \theta_1)} + 1$ $= 2 + e^{i(\theta_1 - \theta_2)} + e^{-i(\theta_1 - \theta_2)} = 2 + 2\cos(\theta_1 - \theta_2) = 2(1 + \cos(\theta_1 - \theta_2))$ Using the identity $1 + \cos(x) = 2\cos^2(x/2)$, $|e^{i\theta_1} + e^{i\theta_2}|^2 = 4\cos^2\left(\frac{\theta_1 - \theta_2}{2}\right)$ $|e^{i\theta_1} + e^{i\theta_2}| = \left|2\cos\left(\frac{\theta_1 - \theta_2}{2}\right)\right| \leq 2$ Explanation: This approach provides a trigonometric verification using polar form and trigonometric identities.

Conclusion for Statement I: Both methods show that $\left|\frac{z_1}{\left|z_1\right|}+\frac{z_2}{\left|z_2\right|}\right| \leq 2$ is always true. Thus, Statement I is correct.

2. Analysis of Statement II

Step 1: Define the common ratio. Let $\frac{a}{|y-z|} = \frac{b}{|z-x|} = \frac{c}{|x-y|} = \lambda$, where $\lambda > 0$. Then $a = \lambda |y-z|$, $b = \lambda |z-x|$, and $c = \lambda |x-y|$. Explanation: Introducing the common ratio helps simplify the given condition.

Step 2: Simplify each term in the sum. Consider $\frac{a^2}{y-z}$. Substituting $a = \lambda |y-z|$, we get $\frac{a^2}{y-z} = \frac{\lambda^2 |y-z|^2}{y-z}$. Since $|y-z|^2 = (y-z)(\overline{y-z}) = (y-z)(\bar{y} - \bar{z})$, we have $\frac{a^2}{y-z} = \frac{\lambda^2 (y-z)(\bar{y} - \bar{z})}{y-z} = \lambda^2 (\bar{y} - \bar{z})$. Similarly, $\frac{b^2}{z-x} = \lambda^2 (\bar{z} - \bar{x})$ and $\frac{c^2}{x-y} = \lambda^2 (\bar{x} - \bar{y})$. Explanation: Using $|w|^2 = w\bar{w}$ allows us to cancel terms and simplify the expressions.

Step 3: Sum the simplified terms. $\frac{a^2}{y-z} + \frac{b^2}{z-x} + \frac{c^2}{x-y} = \lambda^2 (\bar{y} - \bar{z}) + \lambda^2 (\bar{z} - \bar{x}) + \lambda^2 (\bar{x} - \bar{y})$ $= \lambda^2 (\bar{y} - \bar{z} + \bar{z} - \bar{x} + \bar{x} - \bar{y}) = \lambda^2 (0) = 0$. Explanation: The terms telescope and sum to zero.

Conclusion for Statement II: The expression $\frac{a^2}{y-z} + \frac{b^2}{z-x} + \frac{c^2}{x-y}$ evaluates to 0, not 1. Therefore, Statement II is incorrect.

Common Mistakes & Tips

• Remember to use the triangle inequality correctly: $|z_1 + z_2| \leq |z_1| + |z_2|$.
• Don't forget the property $|z|^2 = z\bar{z}$ when dealing with moduli and conjugates.
• Always check for potential simplifications like telescoping sums.

Summary Statement I is correct, as shown by the triangle inequality and polar form representation. Statement II is incorrect, as the given expression simplifies to 0, not 1. Therefore, the correct option is (B).

The final answer is \boxed{B}.