Key Concepts and Formulas

• Modulus of a Complex Number: For a complex number $z = a + bi$, where $a$ and $b$ are real numbers, the modulus is defined as $|z| = \sqrt{a^2 + b^2}$. Also, $|z|^2 = z\bar{z}$, where $\bar{z} = a - bi$ is the complex conjugate of $z$.
• General Equation of a Circle in the Complex Plane: The equation $|z - z_0| = r$ represents a circle with center $z_0$ and radius $r$. The general form is $z\bar{z} + A\bar{z} + \bar{A}z + k = 0$, where the center is $-A$ and the radius is $\sqrt{|A|^2 - k}$.
• Parallelogram Law: For complex numbers $u$ and $v$, $|u + v|^2 + |u - v|^2 = 2(|u|^2 + |v|^2)$.

Step-by-Step Solution

Step 1: Expanding the Given Equation

We are given the equation $|z - \alpha|^2 + |z - \beta|^2 = 2\lambda$. We want to expand this equation using the property $|w|^2 = w\bar{w}$. $|z - \alpha|^2 + |z - \beta|^2 = (z - \alpha)(\bar{z} - \bar{\alpha}) + (z - \beta)(\bar{z} - \bar{\beta}) = 2\lambda$ Expanding the products, we get: $z\bar{z} - z\bar{\alpha} - \bar{z}\alpha + \alpha\bar{\alpha} + z\bar{z} - z\bar{\beta} - \bar{z}\beta + \beta\bar{\beta} = 2\lambda$ Combining like terms: $2z\bar{z} - z(\bar{\alpha} + \bar{\beta}) - \bar{z}(\alpha + \beta) + |\alpha|^2 + |\beta|^2 = 2\lambda$

Step 2: Transforming to Standard Form

To get the equation into the standard form $z\bar{z} + A\bar{z} + \bar{A}z + k = 0$, we divide the entire equation by 2: $z\bar{z} - \frac{z(\bar{\alpha} + \bar{\beta})}{2} - \frac{\bar{z}(\alpha + \beta)}{2} + \frac{|\alpha|^2 + |\beta|^2}{2} = \lambda$ $z\bar{z} - \frac{(\bar{\alpha} + \bar{\beta})}{2}z - \frac{(\alpha + \beta)}{2}\bar{z} + \frac{|\alpha|^2 + |\beta|^2 - 2\lambda}{2} = 0$ Comparing this to the standard form, we identify the center as $C = \frac{\alpha + \beta}{2}$ and the constant term as $k = \frac{|\alpha|^2 + |\beta|^2 - 2\lambda}{2}$. The radius squared is then $R^2 = |A|^2 - k = \left|\frac{\alpha + \beta}{2}\right|^2 - \frac{|\alpha|^2 + |\beta|^2 - 2\lambda}{2}$ $R^2 = \frac{|\alpha + \beta|^2}{4} - \frac{|\alpha|^2 + |\beta|^2 - 2\lambda}{2}$

Step 3: Applying the Parallelogram Law and Solving for $|\alpha - \beta|$

Using the Parallelogram Law, $|\alpha + \beta|^2 = 2(|\alpha|^2 + |\beta|^2) - |\alpha - \beta|^2$. Substituting this into the expression for $R^2$: $R^2 = \frac{2(|\alpha|^2 + |\beta|^2) - |\alpha - \beta|^2}{4} - \frac{|\alpha|^2 + |\beta|^2 - 2\lambda}{2}$ $R^2 = \frac{2(|\alpha|^2 + |\beta|^2) - |\alpha - \beta|^2 - 2(|\alpha|^2 + |\beta|^2 - 2\lambda)}{4}$ $R^2 = \frac{2|\alpha|^2 + 2|\beta|^2 - |\alpha - \beta|^2 - 2|\alpha|^2 - 2|\beta|^2 + 4\lambda}{4}$ $R^2 = \frac{4\lambda - |\alpha - \beta|^2}{4}$ We are given that the radius is $\sqrt{\lambda - 1}$, so $R^2 = \lambda - 1$. Therefore, $\lambda - 1 = \frac{4\lambda - |\alpha - \beta|^2}{4}$ Multiplying both sides by 4 gives: $4\lambda - 4 = 4\lambda - |\alpha - \beta|^2$ $|\alpha - \beta|^2 = 4$ Taking the square root of both sides: $|\alpha - \beta| = 2$

Common Mistakes & Tips

• Incorrect Expansion: Be careful with signs when expanding expressions like $(z-\alpha)(\bar{z}-\bar{\alpha})$. A sign error will propagate through the entire solution.
• Misapplication of Parallelogram Law: Ensure you remember the correct form of the Parallelogram Law.
• Forgetting $|w|^2 = w\bar{w}$: This is the most important identity.

Summary

We expanded the given equation of the circle and transformed it into the standard form. We then used the Parallelogram Law to simplify the expression for the radius squared. Finally, we equated this expression to the given radius squared and solved for $|\alpha - \beta|$, which represents the distance between the complex numbers $\alpha$ and $\beta$. The final result is $|\alpha - \beta| = 2$.

Final Answer The final answer is \boxed{2}, which corresponds to option (A).