JEE Main 2025 April

LEVELJEE Main

Animated Solution for Mathematics - Complex Numbers: Among the statements (S1) : The set ${z \in C - {- i} : ∣ z ∣ = 1 and \frac{z - i}{z + i} is purely real}$ contains exactly two elements, and (S2) : The set ${z \in C - {- 1} : ∣ z ∣ = 1 and \frac{z - 1}{z + 1} is purely imaginary}$ contains infinitely many elements.

Q: Why can't we just use $z = x + iy$ for everything? It feels safer.

You absolutely could, and it would work! However, using $z = x + iy$ is like using a sledgehammer to crack a nut. It leads to messy algebraic expansions that are prone to calculation errors. In complex number problems involving the unit circle, $z = e^{i\theta}$ is the scalpel of complex analysis. It allows you to leverage the symmetry of the circle and the elegance of trigonometric identities, turning a page of algebra into a few clean lines of logic.

Q: How do I know if a complex number is purely imaginary?

A complex number $w$ is purely imaginary if its real part is zero. Algebraically, this is equivalent to $w = -\overline{w}$. In the case of $(S_2)$, we derived the expression $i \tan(\theta/2)$. Since $\tan(\theta/2)$ is a real number for all valid $\theta$, the entire expression is clearly purely imaginary. It is a beautiful result, isn't it?

Select Answer:

Progressive Hints

Hint 1

Hint 1 (The Spark): The unit circle is not just a shape; it is a constraint. When you see

∣ z ∣ = 1

, visualize the complex plane. The conditions 'purely real' and 'purely imaginary' are not just algebraic labels; they are geometric requirements on the argument of the complex number. For

(S_{1})

, ask yourself: what does it mean for a ratio of complex numbers to have no imaginary part?

Hint 2

Hint 2 (The Arsenal): For

(S_{1})

, utilize the property that a complex number

w

is purely real if and only if

w = \overline{w}

. For

(S_{2})

, the Cartesian form

z = x + i y

might be cumbersome. Instead, embrace the power of Euler's formula:

z = e^{i θ}

. This will transform the fraction into a beautiful trigonometric ratio.

Hint 3

Hint 3 (The Execution): For

(S_{1})

, set

\frac{z - i}{z + i} = \overline{(\frac{z - i}{z + i})}

and cross-multiply. Remember that

∣ z ∣ = 1

implies

z \overline{z} = 1

. For

(S_{2})

, substitute

z = e^{i θ}

into

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

and factor out

e^{i θ /2}

from both the numerator and denominator to reveal the tangent function.

Visualized Solution

$I n t r o d u c t i o n t o S t a t e m e n t s S_{1} an d S_{2}$

Statement $(S_{1})$ : $z \in C - {- i}$ , $∣ z ∣ = 1$ , and $\frac{z - i}{z + i}$ is purely real.
Statement $(S_{2})$ : $z \in C - {- 1}$ , $∣ z ∣ = 1$ , and $\frac{z - 1}{z + 1}$ is purely imaginary.

$A na l y z in g S t a t e m e n t S_{1}$

For $(S_{1})$ : $∣ z ∣ = 1 ⟹ z$ lies on the unit circle.
Condition: $z \neq = - i$ .

$P u r e l y R e a l C o n d i t i o n f or S_{1}$

Let $w = \frac{z - i}{z + i}$ .
For $w$ to be purely real, it must equal its conjugate: $w = \overset{w}{ˉ}$ .

$A ppl y in g t h e C o nj ug a t e P r o p er t y$

$\frac{z - i}{z + i} = \frac{z ˉ + i}{z ˉ - i}$
Cross-multiplying: $(z - i) (\overset{z}{ˉ} - i) = (z + i) (\overset{z}{ˉ} + i)$

$S im pl i f y in g t h e E q u a t i o n$

$z \overset{z}{ˉ} - i z - i \overset{z}{ˉ} - 1 = z \overset{z}{ˉ} + i z + i \overset{z}{ˉ} - 1$
Canceling terms: $- i (z + \overset{z}{ˉ}) = i (z + \overset{z}{ˉ})$
$2 i (z + \overset{z}{ˉ}) = 0$

$F in d in g E l e m e n t so f S_{1}$

$z + \overset{z}{ˉ} = 0 ⟹ Re (z) = 0$
Since $∣ z ∣ = 1$ , $z = \pm i$ .
But $z \neq = - i$ , so $z = i$ .
Conclusion: $(S_{1})$ contains exactly one element.

$A na l y z in g S t a t e m e n t S_{2}$

For $(S_{2})$ : $∣ z ∣ = 1$ and $z \neq = - 1$ .
Let $z = e^{i θ}$ where $θ \neq = π$ .

$S u b s t i t u t in g E u l er F or min S_{2}$

We need to check if $f (z) = \frac{z - 1}{z + 1}$ is purely imaginary.
Substitute $z = e^{i θ}$ : $f (z) = \frac{e ^{i θ} - 1}{e ^{i θ} + 1}$

$S im pl i f y in g t h e E x p r ess i o n$

Factor out $e^{i θ /2}$ from numerator and denominator:
$\frac{e ^{i θ /2} ( e ^{i θ /2} - e ^{- i θ /2} )}{e ^{i θ /2} ( e ^{i θ /2} + e ^{- i θ /2} )}$

$C o n v er t in g t o T r i g o n o m e t r i c F or m$

Using $e^{i x} - e^{- i x} = 2 i sin x$ and $e^{i x} + e^{- i x} = 2 cos x$ :
$\frac{2 i s i n ( \frac{θ}{2} )}{2 c o s ( \frac{θ}{2} )} = i tan (\frac{θ}{2})$

$C o n c l u s i o n f or S_{2}$

$i tan (\frac{θ}{2})$ is purely imaginary for all $θ \in [0, 2 π) ∖ {π}$ .
Thus, there are infinitely many such $z$ .

$F ina l A n s w er$

Statement $(S_{1})$ is incorrect (contains 1 element, not 2).
Statement $(S_{2})$ is correct (contains infinitely many elements).
Final Answer: Only (S2) is correct.

00:00 / 00:00

The Sigma Insight: Geometrical Applications of Complex Numbers

Solution Diagram

Analyzing the Setup for $(S_{1})$

We are given the set

{z \in C - {- i} : ∣ z ∣ = 1 and \frac{z - i}{z + i} is purely real}

. The condition

∣ z ∣ = 1

constrains

z

to the unit circle, with the exclusion of the point

z = - i

For a complex number

w = \frac{z - i}{z + i}

to be purely real, it must satisfy the condition

w = \overline{w}

. This leads to the following equation:

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

Solving the Master Equation

By cross-multiplying the equation above, we obtain:

(z - i) (\overline{z} - i) = (z + i) (\overline{z} + i)

Expanding both sides of the equation yields:

z \overline{z} - i z - i \overline{z} - 1 = z \overline{z} + i z + i \overline{z} - 1

The terms

z \overline{z}

and

- 1

cancel out from both sides, simplifying the expression to:

- i (z + \overline{z}) = i (z + \overline{z}) ⟹ 2 i (z + \overline{z}) = 0

Since $2i eq 0$, we must have

z + \overline{z} = 0

, which implies that the real part of

z

is zero. Geometrically,

z

must lie on the imaginary axis.

Conclusion for $(S_{1})$

The intersection of the imaginary axis and the unit circle consists of the points

i

and

- i

. However, the problem explicitly excludes

z = - i

Therefore, the set contains only one element,

z = i

. Since the statement

(S_{1})

claims the set contains two elements, $(S_{1})$ is incorrect.

The Elegance of Euler

Unlocking

(S_{2})

We now examine the set

{z \in C - {- 1} : ∣ z ∣ = 1 and \frac{z - 1}{z + 1} is purely imaginary}

. We represent

z

on the unit circle using the Euler form

z = e^{i θ}

, where $\theta eq \pi$.

Substituting this into the expression, we get:

f (z) = \frac{e ^{i θ} - 1}{e ^{i θ} + 1}

Factoring out

e^{i θ /2}

from the numerator and denominator, we simplify the expression:

\frac{e ^{i θ /2} ( e ^{i θ /2} - e ^{- i θ /2} )}{e ^{i θ /2} ( e ^{i θ /2} + e ^{- i θ /2} )} = \frac{2 i sin ( θ /2 )}{2 cos ( θ /2 )} = i tan (θ /2)

Final Verdict

Since

tan (θ /2)

is a real number for all valid

θ

, the expression

i tan (θ /2)

is always purely imaginary. This condition holds for infinitely many values of

θ

in the interval $

Common Traps & FAQs

Why can't we just use

z = x + i y

for everything? It feels safer.

You absolutely could, and it would work! However, using

z = x + i y

is like using a sledgehammer to crack a nut. It leads to messy algebraic expansions that are prone to calculation errors. In complex number problems involving the unit circle,

z = e^{i θ}

is the scalpel of complex analysis. It allows you to leverage the symmetry of the circle and the elegance of trigonometric identities, turning a page of algebra into a few clean lines of logic.

(S_{1})

, I solved the equation and got

z = i

and

z = - i

. Why is the answer only one element?

This is a classic trap! Always, always check the domain of the problem. The problem explicitly states

z \in C - {- i}

. This means the point

- i

is strictly forbidden. Even if your algebra produces it as a solution, you must reject it because it lies outside the allowed set. Respecting the domain is the hallmark of a mature mathematician.

How do I know if a complex number is purely imaginary?

A complex number

w

is purely imaginary if its real part is zero. Algebraically, this is equivalent to

w = - \overline{w}

. In the case of

(S_{2})

, we derived the expression

i tan (θ /2)

. Since

tan (θ /2)

is a real number for all valid

θ

, the entire expression is clearly purely imaginary. It is a beautiful result, isn't it?

Select Answer:

Visualized Solution

$I n t r o d u c t i o n t o S t a t e m e n t s S_{1} an d S_{2}$

$A na l y z in g S t a t e m e n t S_{1}$

$P u r e l y R e a l C o n d i t i o n f or S_{1}$

$A ppl y in g t h e C o nj ug a t e P r o p er t y$

$S im pl i f y in g t h e E q u a t i o n$

$F in d in g E l e m e n t so f S_{1}$

$A na l y z in g S t a t e m e n t S_{2}$

$S u b s t i t u t in g E u l er F or min S_{2}$

$S im pl i f y in g t h e E x p r ess i o n$

$C o n v er t in g t o T r i g o n o m e t r i c F or m$

$C o n c l u s i o n f or S_{2}$

$F ina l A n s w er$

The Sigma Insight: Geometrical Applications of Complex Numbers

Analyzing the Setup for $(S_{1})$

Solving the Master Equation

Conclusion for $(S_{1})$

The Elegance of Euler

Final Verdict

Similar Questions

Let $A = {z \in C : 1 \leq ∣ z - (1 + i) ∣ \leq 2}$ and $B = {z \in A : ∣ z - (1 - i) ∣ = 1}$ . Then, $B$ :

If $S = {z \in C : \frac{z - i}{z + 2 i} \in R}$ , then :

The number of complex numbers $z$ such that $∣ z - 1∣ = ∣ z + 1∣ = ∣ z - i ∣$ equals

Let $S_{1}, S_{2}$ and $S_{3}$ be three sets defined as $S_{1} = {z \in C : ∣ z - 1∣ \leq 2}$ , $S_{2} = {z \in C : R e ((1 - i) z) \geq 1}$ , $S_{3} = {z \in C : I m (z) \leq 1}$ . Then the set $S_{1} \cap S_{2} \cap S_{3}$

The number of complex numbers z, satisfying $∣ z ∣ = 1$ and $∣ \frac{z}{z} + \frac{z}{z} ∣ = 1$ is:

Let $C$ be the set of all complex numbers. Let $S_{1} = {z \in C : ∣ z - 3 - 2 i ∣^{2} = 8}$ , $S_{2} = {z \in C : Re (z) \geq 5}$ and $S_{3} = {z \in C : ∣ z - \overset{z}{ˉ} ∣ \geq 8}$ . Then the number of elements in $S_{1} \cap S_{2} \cap S_{3}$ is equal to

Let $s, t, r$ be non-zero complex numbers and $L$ be the set of solutions $z = x + i y$ $(x, y \in R, i = - 1)$ of the equation $sz + t \overset{z}{ˉ} + r = 0$ , where $\overset{z}{ˉ} = x - i y$ . Then, which of the following statement(s) is (are) TRUE ?

Comprehension Passage

The number of elements in the set $A \cap B \cap C$ is

Let $z$ be any point in $A \cap B \cap C$ . Then, $∣ z + 1 - i ∣^{2} + ∣ z - 5 - i ∣^{2}$ lies between

Let $z$ be any point $A \cap B \cap C$ and let $w$ be any point satisfying $∣ w - 2 i ∣ < 3$ . Then, $∣ z ∣ - ∣ w ∣ + 3$ lies between

For $n \in N$ , let $S_{n} = {z \in C : ∣ z - (3 + 2 i) ∣ = \frac{n}{4}}$ and $T_{n} = {z \in C : ∣ z - (2 + 3 i) ∣ = \frac{1}{n}}$ . Then the number of elements in the set ${n \in N : S_{n} \cap T_{n} = \emptyset}$ is:

The number of elements in the set ${z = a + ib \in C : a, b \in Z and 1 < ∣ z - 3 + 2 i ∣ < 4}$ is \_\_\_\_\_.

Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } IntroductiontoStatementsS1​andS2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } AnalyzingStatementS1​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } PurelyRealConditionforS1​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ApplyingtheConjugateProperty

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } SimplifyingtheEquation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } FindingElementsofS1​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } AnalyzingStatementS2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } SubstitutingEulerForminS2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } SimplifyingtheExpression

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ConvertingtoTrigonometricForm

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ConclusionforS2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } FinalAnswer

The Sigma Insight: Geometrical Applications of Complex Numbers

Analyzing the Setup for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (S1​)

Solving the Master Equation

Conclusion for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (S1​)

The Elegance of Euler

Final Verdict

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let A={z∈C:1≤∣z−(1+i)∣≤2} and B={z∈A:∣z−(1−i)∣=1}. Then, B :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If S={z∈C:z+2iz−i​∈R}, then :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The number of complex numbers z such that ∣z−1∣=∣z+1∣=∣z−i∣ equals

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The number of complex numbers z, satisfying ∣z∣=1 and ∣zz​+zz​∣=1 is:

Comprehension Passage

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The number of elements in the set A∩B∩C is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let z be any point in A∩B∩C. Then, ∣z+1−i∣2+∣z−5−i∣2 lies between

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let z be any point A∩B∩C and let w be any point satisfying ∣w−2i∣<3. Then, ∣z∣−∣w∣+3 lies between

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } For n∈N, let Sn​={z∈C:∣z−(3+2i)∣=4n​} and Tn​={z∈C:∣z−(2+3i)∣=n1​}. Then the number of elements in the set {n∈N:Sn​∩Tn​=∅} is:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The number of elements in the set {z=a+ib∈C:a,b∈Z and 1<∣z−3+2i∣<4} is \_\_\_\_\_.

$I n t r o d u c t i o n t o S t a t e m e n t s S_{1} an d S_{2}$

$A na l y z in g S t a t e m e n t S_{1}$

$P u r e l y R e a l C o n d i t i o n f or S_{1}$

$A ppl y in g t h e C o nj ug a t e P r o p er t y$

$S im pl i f y in g t h e E q u a t i o n$

$F in d in g E l e m e n t so f S_{1}$

$A na l y z in g S t a t e m e n t S_{2}$

$S u b s t i t u t in g E u l er F or min S_{2}$

$S im pl i f y in g t h e E x p r ess i o n$

$C o n v er t in g t o T r i g o n o m e t r i c F or m$

$C o n c l u s i o n f or S_{2}$

$F ina l A n s w er$

Analyzing the Setup for $(S_{1})$

Conclusion for $(S_{1})$

Let $A = {z \in C : 1 \leq ∣ z - (1 + i) ∣ \leq 2}$ and $B = {z \in A : ∣ z - (1 - i) ∣ = 1}$ . Then, $B$ :

If $S = {z \in C : \frac{z - i}{z + 2 i} \in R}$ , then :

The number of complex numbers $z$ such that $∣ z - 1∣ = ∣ z + 1∣ = ∣ z - i ∣$ equals

The number of complex numbers z, satisfying $∣ z ∣ = 1$ and $∣ \frac{z}{z} + \frac{z}{z} ∣ = 1$ is:

The number of elements in the set $A \cap B \cap C$ is

Let $z$ be any point in $A \cap B \cap C$ . Then, $∣ z + 1 - i ∣^{2} + ∣ z - 5 - i ∣^{2}$ lies between

Let $z$ be any point $A \cap B \cap C$ and let $w$ be any point satisfying $∣ w - 2 i ∣ < 3$ . Then, $∣ z ∣ - ∣ w ∣ + 3$ lies between

For $n \in N$ , let $S_{n} = {z \in C : ∣ z - (3 + 2 i) ∣ = \frac{n}{4}}$ and $T_{n} = {z \in C : ∣ z - (2 + 3 i) ∣ = \frac{1}{n}}$ . Then the number of elements in the set ${n \in N : S_{n} \cap T_{n} = \emptyset}$ is:

The number of elements in the set ${z = a + ib \in C : a, b \in Z and 1 < ∣ z - 3 + 2 i ∣ < 4}$ is \_\_\_\_\_.