Complex Numbers
Complex Number in Iota Form
Complex Numbers_PYQ
Grade 11
Question:
Let $z \in \mathbb{C}$ be such that $|z| < 1$. If $\omega = \dfrac{5 + 3z}{5(1 - z)}$, then
$4\,\text{Im}(\omega) > 5$
$5\,\text{Re}(\omega) > 1$
$5\,\text{Im}(\omega) < 1$
$5\,\text{Re}(\omega) > 4$
Step-by-Step Solution
Key Concept: Transform $\omega$ into $c + \frac{k}{1-z}$ form, then convert the inequality on $\text{Re}(\omega)$ back to a condition on $|z|$.
**Step 1: Decompose ω**
Write $5 + 3z = -3(1-z) + 8$, so $\omega = \dfrac{-3(1-z)+8}{5(1-z)} = -\dfrac{3}{5} + \dfrac{8}{5(1-z)}$.
**Step 2: Find Re(ω)**
Let $z = x+iy$. Then $\text{Re}\!\left(\dfrac{1}{1-z}\right) = \dfrac{1-x}{(1-x)^2 + y^2}$. So $\text{Re}(\omega) = -\dfrac{3}{5} + \dfrac{8(1-x)}{5[(1-x)^2+y^2]}$.
**Step 3: Show 5 Re(ω) > 1 iff |z| < 1**
$5\,\text{Re}(\omega) > 1 \iff \dfrac{8(1-x)}{(1-x)^2+y^2} > 4 \iff 2(1-x) > (1-x)^2 + y^2 \iff 0 > (1-x)^2 - 2(1-x) + y^2 = x^2 + y^2 - 1$, which is exactly $|z| < 1$. ✓
Correct Answer: 2