Key Concepts and Formulas

• A complex number $w$ is purely imaginary if $w + \overline{w} = 0$.
• For any complex number $z$, $z\overline{z} = |z|^2$.
• If $\alpha$ is a real number, then $\overline{\alpha} = \alpha$.

Step-by-Step Solution

Step 1: Define the complex number and apply the purely imaginary condition

Let $w = \frac{z - \alpha}{z + \alpha}$. Since $w$ is purely imaginary, we have $w + \overline{w} = 0$. This means $\frac{z - \alpha}{z + \alpha} + \overline{\left(\frac{z - \alpha}{z + \alpha}\right)} = 0$ This step is crucial as it directly uses the defining property of a purely imaginary number to set up an equation.

Step 2: Simplify the conjugate

Using the properties of conjugates, we can simplify $\overline{\left(\frac{z - \alpha}{z + \alpha}\right)}$ as follows: $\overline{\left(\frac{z - \alpha}{z + \alpha}\right)} = \frac{\overline{z - \alpha}}{\overline{z + \alpha}} = \frac{\overline{z} - \overline{\alpha}}{\overline{z} + \overline{\alpha}}$ Since $\alpha$ is real, $\overline{\alpha} = \alpha$. Thus, $\overline{w} = \frac{\overline{z} - \alpha}{\overline{z} + \alpha}$ Now, substitute this back into the equation from Step 1: $\frac{z - \alpha}{z + \alpha} + \frac{\overline{z} - \alpha}{\overline{z} + \alpha} = 0$ This simplifies the equation by expressing the conjugate in terms of $\overline{z}$ and $\alpha$.

Step 3: Combine the fractions

To solve the equation, we combine the fractions by finding a common denominator: $\frac{(z - \alpha)(\overline{z} + \alpha) + (\overline{z} - \alpha)(z + \alpha)}{(z + \alpha)(\overline{z} + \alpha)} = 0$ For this fraction to be zero, the numerator must be zero: $(z - \alpha)(\overline{z} + \alpha) + (\overline{z} - \alpha)(z + \alpha) = 0$ Multiplying out the terms: $(z\overline{z} + z\alpha - \alpha\overline{z} - \alpha^2) + (\overline{z}z + \overline{z}\alpha - \alpha z - \alpha^2) = 0$ This step removes the denominator and prepares the equation for simplification.

Step 4: Simplify the expression

Now, we simplify the expression. Recall that $z\overline{z} = |z|^2$, and since $\alpha$ is real, $z\alpha = \alpha z$ and $\overline{z}\alpha = \alpha\overline{z}$. $|z|^2 + z\alpha - \alpha\overline{z} - \alpha^2 + |z|^2 + \overline{z}\alpha - \alpha z - \alpha^2 = 0$ $2|z|^2 - 2\alpha^2 = 0$ $|z|^2 - \alpha^2 = 0$ Thus, we have $|z|^2 = \alpha^2$ This simplification isolates $|z|^2$ and $\alpha^2$, making it easy to substitute the given value of $|z|$.

Step 5: Substitute the value of |z| and solve for α

We are given that $|z| = 2$. Substituting this into the equation $|z|^2 = \alpha^2$, we get: $(2)^2 = \alpha^2$ $4 = \alpha^2$ $\alpha = \pm 2$ Therefore, a possible value of $\alpha$ is $2$.

Common Mistakes & Tips

• Remember that $\overline{\alpha} = \alpha$ only if $\alpha$ is real. This is crucial for simplifying the conjugate.
• Be careful with signs when expanding the products in step 3.
• Recall the definition of modulus: $|z|^2 = z\overline{z}$.

Summary

We started with the given condition that $\frac{z-\alpha}{z+\alpha}$ is purely imaginary, meaning $w + \overline{w} = 0$. We simplified the conjugate and substituted it back into the equation. Through algebraic manipulation and using the fact that $\alpha$ is real, we arrived at the equation $|z|^2 = \alpha^2$. Finally, we used the given value of $|z| = 2$ to find that $\alpha = \pm 2$. Therefore, a value of $\alpha$ is 2.

Final Answer

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