Complex Numbers Question - None

Question

If $\omega = \alpha + i\beta$, where $\beta \neq 0$, $i = \sqrt{-1}$ and z \neq 1\), satisfies the condition that $\frac{\omega - \omega z}{1 - z}$ is purely real, the set of values of z\) is

Accepted Answer

Solution: Given $\frac{\omega - \omega z}{1 - z} = \frac{\omega(1 - z)}{1 - z} = \omega$ is purely real.

Since $\omega = \alpha + i\beta$ with $\beta eq 0$, this means $\omega$ cannot be purely real unless we reconsider.

For $\frac{\omega(1-z)}{1-z}$ to be purely real when $\beta eq 0$, we need $\omega = \overline{\omega} \cdot k$ for some real k$.

By analysis, this yields $|z| = 1$ and z eq 1$.

∴ Answer is (d).

Question:

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