Complex Numbers
Complex Number in Iota Form
Complex Numbers_PYQ
Grade 11
Question:
All the points in the set $S = \left\{\dfrac{\alpha + i}{\alpha - i} : \alpha \in \mathbb{R}\right\}$ $(i = \sqrt{-1})$ lie on a
circle whose radius is $\sqrt{2}$
straight line whose slope is $-1$
circle whose radius is $1$
straight line whose slope is $1$
Step-by-Step Solution
Key Concept: For $z = \frac{a}{\bar{a}}$ where $a$ and $\bar{a}$ are complex conjugates, $|z| = 1$ always. Note that $\alpha - i = \overline{\alpha + i}$ for real $\alpha$.
**Step 1: Compute the modulus**
Let $z = \dfrac{\alpha + i}{\alpha - i}$. Then $|z| = \dfrac{|\alpha + i|}{|\alpha - i|} = \dfrac{\sqrt{\alpha^2+1}}{\sqrt{\alpha^2+1}} = 1$ for all $\alpha \in \mathbb{R}$.
**Step 2: Conclude the locus**
Since $|z| = 1$ for every $\alpha$, all points lie on the unit circle, which has radius $1$.
Correct Answer: 3