Complex Numbers Question - None

Question

The set of all $\alpha \in R$, for which $w = \dfrac{1+(1-8\alpha)z}{1-z}$ is a purely imaginary number, for all $z \in C$ satisfying $|z|=1$ and $\text{Re } z \neq 1$, is

Accepted Answer

Step 1: Let z = x + iy where |z|=1, so x² + y² = 1 with x ≠ 1.

Step 2: Compute w = (1+(1-8α)z)/(1-z). For w to be purely imaginary, Re(w) = 0.

Step 3: Write w = (1+(1-8α)(x+iy))/((1-x)-iy). Multiply numerator and denominator by the conjugate (1-x)+iy:

w = [(1+(1-8α)x + i(1-8α)y)·((1-x)+iy)] / [(1-x)² + y²]

Step 4: The denominator is (1-x)² + y² = 1 - 2x + x² + y² = 2 - 2x = 2(1-x).

Step 5: Expand the numerator's real part:
Re = [1+(1-8α)x](1-x) - (1-8α)y²
= (1-x) + (1-8α)x(1-x) - (1-8α)(1-x²)
= (1-x)[1 + (1-8α)x - (1-8α)(1+x)]
= (1-x)[1 + (1-8α)(x-1-x)]
= (1-x)[1 - (1-8α)]
= (1-x)·8α

Step 6: For Re(w) = 0 for all z with |z|=1 and x ≠ 1, we need 8α = 0 (since 1-x ≠ 0).

Step 7: Therefore α = 0.

∴ Answer: {0}

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