All the Possible Points in the Set S = {(α + i)/(α − i) : α ∈ R, i = √−1} Lie On

JEE Mains 2019 — Previous Year Question.

Problem

All the possible points in the set $S=\left\{\dfrac{\alpha+i}{\alpha-i}:\alpha\in\mathbb R,\ i=\sqrt{-1}\right\}$ lie on:

Key insight. Multiplying numerator and denominator by the conjugate of the denominator turns the quotient into a single expression with a real denominator — this exposes $x$ and $y$ as explicit functions of $\alpha$, and a well-known algebraic identity then eliminates $\alpha$ in one step.

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Approach

Rationalize $\dfrac{\alpha+i}{\alpha-i}$ by multiplying top and bottom by $\alpha+i$ (the conjugate of the denominator), which turns the denominator into the real number $\alpha^2+1$. Writing the result as $x+iy$ gives $x$ and $y$ each as explicit rational functions of $\alpha$. The identity $(\alpha^2+1)^2=(\alpha^2-1)^2+4\alpha^2$ then shows that $x^2+y^2$ collapses to exactly $1$, regardless of $\alpha$.

Solution

Step 1 — Rationalize the expression

$$z = \frac{\alpha+i}{\alpha-i} \times \frac{\alpha+i}{\alpha+i} = \frac{(\alpha+i)^2}{\alpha^2+1}$$

Step 2 — Expand the numerator

$$(\alpha+i)^2 = \alpha^2+2\alpha i+i^2 = (\alpha^2-1)+2\alpha i$$

So:

$$z = \frac{(\alpha^2-1)+2\alpha i}{\alpha^2+1} = \frac{\alpha^2-1}{\alpha^2+1}+i\cdot\frac{2\alpha}{\alpha^2+1}$$

Step 3 — Identify x and y

Writing $z=x+iy$:

$$x = \frac{\alpha^2-1}{\alpha^2+1}, \qquad y = \frac{2\alpha}{\alpha^2+1}$$

Step 4 — Compute x² + y²

$$x^2+y^2 = \frac{(\alpha^2-1)^2+(2\alpha)^2}{(\alpha^2+1)^2} = \frac{(\alpha^2-1)^2+4\alpha^2}{(\alpha^2+1)^2}$$

Step 5 — Simplify the numerator using an algebraic identity

Expanding the numerator:

$$(\alpha^2-1)^2+4\alpha^2 = \alpha^4-2\alpha^2+1+4\alpha^2 = \alpha^4+2\alpha^2+1 = (\alpha^2+1)^2$$

Step 6 — Substitute back

$$x^2+y^2 = \frac{(\alpha^2+1)^2}{(\alpha^2+1)^2} = 1$$

This holds for every real value of $\alpha$ — the parameter has cancelled out completely.

Answer

Common mistakes

• Forgetting to multiply by the conjugate. Working directly with $\dfrac{\alpha+i}{\alpha-i}$ leaves a complex number in the denominator, from which real and imaginary parts can’t be read off directly — rationalizing first, as in Step 1, is what makes $x$ and $y$ explicit.
• Missing the identity $(\alpha^2-1)^2+4\alpha^2=(\alpha^2+1)^2$. Without spotting this, the numerator in Step 4 looks like a fourth-degree mess in $\alpha$ that doesn’t obviously simplify — recognising it as $(\alpha^2+1)^2$ in disguise is the one algebraic step that makes the whole problem collapse to a clean answer.
• Assuming the answer must depend on which specific $\alpha$ is chosen. Since $\alpha$ genuinely cancels out of $x^2+y^2$ entirely, every point of $S$ lies on the same circle — there’s no need to check multiple values of $\alpha$ to be sure, though doing so (e.g. $\alpha=0$ gives $z=-1$, and indeed $(-1)^2+0^2=1$) is a quick sanity check.

Practise next

• Find what curve the points of $S=\left\{\dfrac{\alpha+2i}{\alpha-2i}:\alpha\in\mathbb R\right\}$ lie on.