Question

The set of all $\alpha \epsilon R$, for which $w = \frac{1 + (1 - 8 \alpha)z}{1 - z}$ is a purely imaginary number, for all $z \neq 1$, is :

• A. an empty set
• B. $\{ 0 \}$
• C. $\left\{0 , \frac{1}{4} , - \frac{1}{4} \right\}$
• D. equal to R

Answer 1

The Correct Option is B

To determine the set of all real numbers \alpha for which the expression w = \frac{1 + (1 - 8\alpha)z}{1 - z} is purely imaginary for all z \neq 1, we need to look at the real and imaginary parts of w and set the real part to zero.

1. w is purely imaginary if its real part is zero. Expressing w in terms of z gives:
   w = \frac{1 + (1 - 8\alpha)z}{1 - z} = \frac{1 + z - 8\alpha z}{1 - z}
2. Re-write w with real and imaginary components. Let z = x + yi, where x and y are real numbers and i is the imaginary unit:
   w = \frac{1 + (1 - 8\alpha)(x + yi)}{1 - (x + yi)}
3. To find when this is purely imaginary, simplify and equate the real part of w to zero:
   1 + (1 - 8\alpha)x = 0
   By solving, we have:
   1 + x - 8\alpha x = 0 \quad \Rightarrow \quad 8\alpha x = 1 + x
4. The equation 8\alpha x = 1 + x holds for all x only if both sides are zero. Thus, setting x = 0 gives:
   1 + x = 0 \Rightarrow x = -1,
5. However, for w to be purely imaginary at this value, the term involving \alpha must also equal 1 for x \neq 0. Hence: \alpha must be set such that other conditions do not introduce real terms, concluding only:
   \alpha = 0.

Thus the set of \alpha for which w is purely imaginary for all permissible z is \{ 0 \}.