Among the statements (S1): The set z C - -i:|z| = 1 and (z - i/z + i) is purely real contains exactly two elements, and (S2) : The set z C - -1 : |z| = 1 and (z - 1/z + 1) is purely imaginary contains infinitely many elements.

A complex number w is purely real if w = w. A complex number w is purely imaginary if w + w = 0. Let's evaluate statement **(S1)**: If w is purely real, then w = w: Since z + z = 2Re(z) = 0, z must lie on the imaginary axis (y-axis). Given the condition |z| = 1, the only points are z = i and z = -i. However, the domain excludes z = -i. Let's test z = i: For z = i, (i - i/i + i) = 0, which is purely real. So it contains elements on the unit circle. But the condition z + z = 0 alongside |z|=1 explicitly limits it to z=i only, which is one element, not two. Thus, (S1) is incorrect. Let's evaluate statement **(S2)**: If u is purely imaginary, then u + u = 0: This condition holds true for ALL points on the unit circle |z| = 1 except z = -1 (which makes the denominator zero). Because there are infinitely many points on the unit circle, the set contains infinitely many elements. Thus, (S2) is correct. Geometric shortcut: The transformation w = (z-1/z+1) maps the unit circle |z|=1 directly onto the imaginary axis Re(w)=0. Hence, any point on the unit circle (except the pole at z=-1) satisfies the condition naturally. Class 11 Mathematics: Complex Numbers and Quadratic Equations

Among the statements (S1): The set z C - -i:|z| = 1 and (z - i/z + i) is purely real contains exactly two elements, and (S2) : The set z C - -1 : |z| = 1 and (z - 1/z + 1) is purely imaginary contains infinitely many elements.

A complex number w is purely real if w = w. A complex number w is purely imaginary if w + w = 0. Let's evaluate statement **(S1)**: If w is purely real, then w = w: Since z + z = 2Re(z) = 0, z must lie on the imaginary axis (y-axis). Given the condition |z| = 1, the only points are z = i and z = -i. However, the domain excludes z = -i. Let's test z = i: For z = i, (i - i/i + i) = 0, which is purely real. So it contains elements on the unit circle. But the condition z + z = 0 alongside |z|=1 explicitly limits it to z=i only, which is one element, not two. Thus, (S1) is incorrect. Let's evaluate statement **(S2)**: If u is purely imaginary, then u + u = 0: This condition holds true for ALL points on the unit circle |z| = 1 except z = -1 (which makes the denominator zero). Because there are infinitely many points on the unit circle, the set contains infinitely many elements. Thus, (S2) is correct. Geometric shortcut: The transformation w = (z-1/z+1) maps the unit circle |z|=1 directly onto the imaginary axis Re(w)=0. Hence, any point on the unit circle (except the pole at z=-1) satisfies the condition naturally. Class 11 Mathematics: Complex Numbers and Quadratic Equations

Among the statements (S1): The set z C - -i:|z| = 1 and (z - i/z + i) is purely real contains exactly two elements, and…

Solution & Explanation

### Related Formula A complex number

w

is purely real if

w = barw

. A complex number

w

is purely imaginary if

w + barw = 0

. ### Core Logic Let's evaluate statement **(S1)**:

w = fracz - iz + i

w

is purely real, then

w = barw

fracz - iz + i = fracbarz + ibarz - i

(z - i)(barz - i) = (z + i)(barz + i)

|z|^2 - iz - ibarz - 1 = |z|^2 + iz + ibarz - 1

-i(z + barz) = i(z + barz) implies 2i(z + barz) = 0 implies z + barz = 0

Since

z + barz = 2textRe(z) = 0

z

must lie on the imaginary axis (y-axis). Given the condition

|z| = 1

, the only points are

z = i

and

z = -i

. However, the domain excludes

z = -i

. Let's test

z = i

: For

z = i

fraci - ii + i = 0

, which is purely real. So it contains elements on the unit circle. But the condition

z + barz = 0

alongside

|z|=1

explicitly limits it to

z=i

only, which is one element, not two. Thus, (S1) is incorrect. ### Step 1: Evaluate Statement S2 Let's evaluate statement **(S2)**:

u = fracz - 1z + 1

u

is purely imaginary, then

u + baru = 0

fracz - 1z + 1 + fracbarz - 1barz + 1 = 0

frac(z - 1)(barz + 1) + (z + 1)(barz - 1)(z + 1)(barz + 1) = 0

(|z|^2 + z - barz - 1) + (|z|^2 - z + barz - 1) = 0

2|z|^2 - 2 = 0 implies |z|^2 = 1 implies |z| = 1

This condition holds true for ALL points on the unit circle

|z| = 1

except

z = -1

(which makes the denominator zero). Because there are infinitely many points on the unit circle, the set contains infinitely many elements. Thus, (S2) is correct. ### Pattern Recognition Geometric shortcut: The transformation

w = fracz-1z+1

maps the unit circle

|z|=1

directly onto the imaginary axis

textRe(w)=0

. Hence, any point on the unit circle (except the pole at

z=-1

) satisfies the condition naturally. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Complex Numbers and Quadratic Equations

More Complex Numbers Previous-Year Questions — Page 7

Q26 jee_main_2024_31_jan_morning Properties of Modulus and Argument

alpha

denotes the number of solutions of

|1 - i|^x = 2^x

and

beta = left(frac|z|arg(z)right)

, where

z = fracpi4 (1 + i)^4 left(frac1 - sqrtpi isqrtpi + i + fracsqrtpi - i1 + sqrtpi iright), i = sqrt-1

, then the distance of the point

(alpha, beta)

from the line

4x - 3y = 7

Numerical Answer. Answer: 3 to 3

Solution

### Core Logic

|1 - i|^x = 2^x implies (sqrt2)^x = 2^x implies 2^x/2 = 2^x

This implies

fracx2 = x implies x = 0

. There is exactly 1 solution, so

alpha = 1

. ### Step 1: Simplify complex number z

(1+i)^4 = ((1+i)^2)^2 = (1 + i^2 + 2i)^2 = (2i)^2 = -4

Thus,

z = -pi left( frac(1-sqrtpii)(sqrtpi-i)pi + 1 + frac(sqrtpi-i)(1-sqrtpii)1 + pi right)

### Step 2: Simplify Bracket Let's expand the terms directly:

z = fracpi4(-4) left[ fracsqrtpi - pi i - i - sqrtpipi + 1 + fracsqrtpi - i - pi i - sqrtpi1 + pi right]

= -pi left[ frac-i(pi+1)pi+1 + frac-i(pi+1)pi+1 right]

= -pi [ -i - i ] = 2pi i

### Step 3: Find beta For

z = 2pi i

|z| = 2pi

and

arg(z) = fracpi2

beta = frac|z|arg(z) = frac2pipi/2 = 4

### Step 4: Distance from Line Distance of point

(alpha, beta) = (1, 4)

from the line

4x - 3y - 7 = 0

D = frac|4(1) - 3(4) - 7|sqrt4^2 + (-3)^2 = frac|4 - 12 - 7|5 = frac|-15|5 = 3

### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Complex Numbers and Quadratic Equations Class 11 Maths: Straight Lines

Among the statements (S1): The set $\zin mathbbC - \-i\:|z| = 1text and fracz - iz + itext is purely real\$ contains exactly two elements, and (S2) : The set $\z in mathbbC - \-1\ : |z| = 1text and fracz - 1z + 1text is purely imaginary\$ contains infinitely many elements.

Solution & Explanation

Reference Study Guides

More Complex Numbers Previous-Year Questions — Page 7

Solution

More Complex Numbers Questions — jee_main_2025_07_april_morning

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Solution & Explanation

Reference Study Guides

More Complex Numbers Previous-Year Questions — Page 7

Solution

More Complex Numbers Questions — jee_main_2025_07_april_morning

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