Complex Numbers on Unit Circle Purely Real Imaginary

 

❓ Question
Among the following statements:
(S1)
$$\{z\in \mathbb{C}\setminus \{-i\}:\mathrm{\mid }z\mathrm{\mid }=1\text{ and }\frac{z-i}{z+i}\text{ is purely real}\}<br>$$contains exactly two elements.
(S2)
$$\{z\in \mathbb{C}\setminus \{-1\}:\mathrm{\mid }z\mathrm{\mid }=1\text{ and }\frac{z-1}{z+1}\text{ is purely imaginary}\}<br>$$contains infinitely many elements.
Determine which statement(s) is/are true.

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🖼️ Question Image

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✍️ Short Explanation

This problem is based on:

👉 Complex numbers on unit circle
👉 Purely real / purely imaginary conditions
👉 Algebraic simplification.

Main idea:

Use:

$$|z|=1 \Rightarrow z=\cos\theta+i\sin\theta$$

or directly use conjugate properties.

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🔷 Step 1 — Check (S1) 💯

Given:

$$|z|=1$$

Let:

$$z=e^{i\theta}$$

We need:

$$\frac{z-i}{z+i}$$

to be purely real.

Substitute:

$$z=\cos\theta+i\sin\theta$$

Then:

$$\frac{z-i}{z+i} = \frac{\cos\theta+i(\sin\theta-1)} {\cos\theta+i(\sin\theta+1)}$$

Multiply numerator and denominator by conjugate of denominator.

Imaginary part becomes zero when:

$$\cos\theta=0$$

Thus:

$$\theta=\frac\pi2,\frac{3\pi}2$$

giving:

$$z=i,-i$$

But:

$$z\ne-i$$

Hence only:

$$\boxed{ z=i }$$

satisfies.

So set contains exactly one element, not two.

❌ (S1) is incorrect.

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🔷 Step 2 — Check (S2)

Need:

$$\frac{z-1}{z+1}$$

purely imaginary with:

$$|z|=1$$

Take:

$$z=e^{i\theta}$$

Then:

$$\frac{z-1}{z+1} = \frac{e^{i\theta}-1}{e^{i\theta}+1}$$

Using standard identity:

$$\boxed{ \frac{e^{i\theta}-1}{e^{i\theta}+1} = i\tan\frac\theta2 }$$

which is purely imaginary for all admissible $\theta$.

Only restriction:

$$z\ne-1$$

So infinitely many points on unit circle satisfy.

✔ (S2) is correct.

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🔷 Step 3 — Final Conclusion

$$(S1)\text{ is false}$$
$$(S2)\text{ is true}$$

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✅ Final Answer

$$\boxed{ \text{Only (S2) is correct} }$$

(Option 1)

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🔷 JEE Trap Alert 🚨

❌ Forgetting exclusion:

$$z\ne -i,\quad z\ne -1$$

❌ Not using unit-circle identity:

$$\boxed{{\displaystyle \frac{e^{i\theta }-1}{e^{i\theta }+1}=i\tan \frac{\theta }{2}}}$$

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