JEE Main: How Many Solutions on |z| = 1? Complex Number Trick 💡

 

❓ Question

Among the following statements:

(S1)

$$\{z\in \mathbb{C}\setminus \{-i\}:\mathrm{\mid }z\mathrm{\mid }=1\text{ and }\frac{z-i}{z+i}\text{ is purely real}\}$$

contains exactly two elements.

(S2)

$$\{z\in \mathbb{C}\setminus \{-1\}:\mathrm{\mid }z\mathrm{\mid }=1\text{ and }\frac{z-1}{z+1}\text{ is purely imaginary}\}$$

contains infinitely many elements.

Determine which statement(s) is/are true.

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🖼️ Question Image

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✍️ Short Solution

This question is based on a very powerful JEE concept:

For a complex number

$$\frac{z-a}{z-b}$$

• Purely real ⇔ arguments are equal or differ by π

• Purely imaginary ⇔ arguments differ by $\frac{\pi}{2}$

We’ll use argument geometry on the unit circle.

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🔹 Key Concept (Must Know)

For any non-zero complex number $w$:

• $w$ is purely real ⇔

  $$\arg (w)=0\text{ or }\pi$$

• $w$ is purely imaginary ⇔

  $$\arg (w)=\pm \frac{\pi }{2}$$

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🔍 Statement (S1) Analysis

Condition:

$$|z| = 1,\quad \frac{z - i}{z + i} \in \mathbb{R}$$

Step 1 — Convert to argument form

$$\arg\left(\frac{z - i}{z + i}\right) = \arg(z - i) - \arg(z + i)$$

For this to be purely real:

$$\arg(z - i) - \arg(z + i) = 0 \text{ or } \pi$$

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Step 2 — Geometrical interpretation

• $z$ lies on the unit circle

• Fixed points: $i$ and $-i$
  

• Condition means:
  👉 The lines joining $z$ to $i$ and $-i$ make either:

  • same direction

  • or opposite direction

This happens only at two points on the unit circle:

$$z = 1 \quad \text{and} \quad z = -1$$

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✔ Conclusion for (S1)

Exactly two complex numbers satisfy the condition.

$$\boxed{\text{(S1) is TRUE}}$$

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🔍 Statement (S2) Analysis

Condition:

$$|z| = 1,\quad \frac{z - 1}{z + 1} \text{ is purely imaginary}$$

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Step 1 — Argument condition

Purely imaginary ⇒

$$\arg(z - 1) - \arg(z + 1) = \pm \frac{\pi}{2}$$

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Step 2 — Geometrical meaning

This means:

• The angle between vectors $(z - 1)$ and $(z + 1)$ is 90°

• Points 1 and −1 are fixed

• $z$ moves on the unit circle

👉 The locus of points forming a right angle with fixed endpoints is a circle.

But here, the unit circle intersects this locus at infinitely many points.

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✔ Conclusion for (S2)

There are infinitely many points on $|z| = 1$ satisfying the condition.

$$\boxed{\text{(S2) is TRUE}}$$

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✅ Final Answer

$$\boxed{\text{Both (S1) and (S2) are TRUE}}$$