Complex Numbers
Purely Real Condition via Conjugate
Complex Numbers_PYQ
Grade 11
Question:
If $w=\alpha+i\beta$, where $\beta\neq0$ and $z\neq1$, satisfies the condition that $\left(\dfrac{w-\bar{w}z}{1-z}\right)$ is purely real, then the set of values of $z$ is
$\{z:|z|=1,\;z\neq2\}$
$\{z:|z|=1,\;z\neq1\}$
$\{z:z=\bar{z}\}$
None of these
Step-by-Step Solution
Key Concept: The imaginary part of $\dfrac{w-\bar{w}z}{1-z}$ evaluates cleanly to $\dfrac{\beta(1-|z|^2)}{|1-z|^2}$. Setting this to zero with $\beta\neq0$ directly gives $|z|=1$.
**Step 1: Compute numerator**
With $w=\alpha+i\beta$, $\bar{w}=\alpha-i\beta$, $z=x+iy$: $w-\bar{w}z=(\alpha(1-x)-\beta y)+i(\beta(1+x)-\alpha y)$.
**Step 2: Divide by 1−z and extract imaginary part**
After multiplying by the conjugate of $1-z=(1-x)-iy$, the imaginary part of $\dfrac{w-\bar{w}z}{1-z}$ simplifies to $\dfrac{\beta(1-x^2-y^2)}{(1-x)^2+y^2}=\dfrac{\beta(1-|z|^2)}{|1-z|^2}$.
**Step 3: Set imaginary part to zero**
$\beta(1-|z|^2)=0$. Since $\beta\neq0$ and $|1-z|^2\neq0$ (as $z\neq1$), we get $|z|=1$.
Correct Answer: 2