Let $P=\zinmathbbC:|z+2-3i|le1\$ $P=\{z\in\mathbb{C}:|z+2-3i|\le1\}$ and $Q=\zinmathbbC:z(1+i)+overlinez(1-i)le-8\$ . Let in $P cap Q, |z-3+2i|$ $P \cap Q, |z-3+2i|$ be maximum and minimum at $z_1$ $z_{1}$ and $z_2$ $z_{2}$ respectively. If $|z_1|^2+2|z_2|^2=alpha+betasqrt2$ , where $alpha, beta$ $\alpha, \beta$ are integers, then $alpha+beta$ $\alpha+\beta$ equals

Numerical Answer Type:

Enter a numerical value Answer: 36 to 36 +4 marks

Solution & Explanation

### Related Formula For a complex coordinate transformation, substituting

z = x + iy

and its conjugate

overlinez = x - iy

maps a complex condition directly into rectangular Cartesian coordinates. ### Core Logic Let's translate the complex set properties into Cartesian geometry: - Set

P

|z - (-2 + 3i)| le 1 implies

Interior and boundary of a circle with center

C(-2, 3)

and radius

r = 1

. - Set

Q

(x+iy)(1+i) + (x-iy)(1-i) le -8 implies (x - y + ix + iy) + (x - y - ix - iy) le -8

2(x - y) le -8 implies x - y + 4 le 0

This defines a half-plane below or to the left of the boundary line

L_2: x - y + 4 = 0

. ### Step 1: Identify Extreme Points for Distance from Point A We want to find points in the region

P cap Q

that minimize and maximize the distance to the external point

A(3, -2)

, corresponding to

|z - (3 - 2i)|

. The line

L_1

connecting center

C(-2, 3)

and point

A(3, -2)

has slope:

m = frac-2 - 33 - (-2) = frac-55 = -1

Equation of line

L_1

y - 3 = -1(x + 2) implies x + y - 1 = 0

Geometric region intersection for complex inequalities for Q27 - JEE Main 2024 01 February Morning — The graphic details the intersection region bounded by the circular locus and the line inequality, showing points z1 and z2 relative to external reference point P.

### Step 2: Calculate Coordinates for z1 and z2 - **Minimum Distance Point (

z_2

):** By geometric observation, the minimum distance from

A

to the bounded region is the intersection point of lines

L_1

and

L_2

x - y + 4 = 0 quad textand quad x + y - 1 = 0 implies 2x + 3 = 0 implies x = -frac32, \, y = frac52

So,

z_2 = left(-frac32, frac52 ight)

. - **Maximum Distance Point (

z_1

):** The maximum distance is at the far boundary edge of the circle along line

L_1

. The vector path from

C(-2, 3)

opposite to

A

has unit direction

left(-frac1sqrt2, frac1sqrt2right)

z_1 = left(-2 - frac1sqrt2, \, 3 + frac1sqrt2right)

### Step 3: Evaluate Magnitudes and Sum Parameters Calculate the squares of the moduli:

|z_1|^2 = left(-2 - frac1sqrt2right)^2 + left(3 + frac1sqrt2right)^2 = 4 + 2sqrt2 + frac12 + 9 + 3sqrt2 + frac12 = 14 + 5sqrt2

|z_2|^2 = left(-frac32right)^2 + left(frac52 ight)^2 = frac94 + frac254 = frac344 = frac172

Now compute the total requested term:

|z_1|^2 + 2|z_2|^2 = (14 + 5sqrt2) + 2left(frac172right) = 14 + 5sqrt2 + 17 = 31 + 5sqrt2

Matching with

alpha + betasqrt2

gives

alpha = 31

and

beta = 5

. Thus:

alpha + beta = 31 + 5 = 36

### Pattern Recognition Sees: Locus intersection involving geometric complex inequalities. Shortcut: Translating complex equations into standard 2D graphs reveals the geometry instantly, mapping extreme distances to line intersections or boundary nodes cleanly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Complex Numbers: Geometry Class 11 Coordinate Geometry: Straight Lines

Reference Study Guides

Complex Numbers

More Complex Numbers Previous-Year Questions

Q75 jee_main_2025_02_april_evening Quadratic Equations

If the set of all

a in mathbbR - \1\

, for which the roots of the equation

(1 - a)x^2 + 2(a - 3)x + 9 = 0

are positive is

(-infty, -alpha] cup [beta, gamma)

, then

2alpha + beta + gamma

is equal to ____________.

Numerical Answer. Answer: 7 to 7

Solution

### Related Formula

textFor a quadratic equation A x^2 + B x + C = 0 text to have two positive real roots:

text1. Real roots: D = B^2 - 4AC ge 0

text2. Sum of roots: -fracBA > 0

text3. Product of roots: fracCA > 0

### Core Logic We write down three systems of inequalities based on real and positive root conditions, find their intersection, and map the boundaries to solve for the parameters. ### Step 1: Apply the discriminant condition (Real roots) For real roots, the discriminant

D ge 0

D = left[ 2(a - 3) right]^2 - 4(1 - a)(9) ge 0

4(a^2 - 6a + 9) - 36(1 - a) ge 0

(a^2 - 6a + 9) - 9(1 - a) ge 0

a^2 - 6a + 9 - 9 + 9 a ge 0

a^2 + 3a ge 0 implies a(a + 3) ge 0

Thus, the interval is:

a in (-infty, -3] cup [0, infty) quad text--- (1)

### Step 2: Apply the sum of roots condition (Positive sum) For positive roots, the sum of roots must be positive:

-fracBA = frac-2(a - 3)1 - a = frac2(a - 3)a - 1 > 0

Using the wavy curve method for

fraca-3a-1 > 0

a in (-infty, 1) cup (3, infty) quad text--- (2)

### Step 3: Apply the product of roots condition (Positive product) For positive roots, the product of roots must be positive:

fracCA = frac91 - a > 0 implies 1 - a > 0 implies a < 1

Thus, the interval is:

a in (-infty, 1) quad text--- (3)

### Step 4: Find the intersection of all conditions Intersecting equations (1), (2), and (3): - First, intersect (2) and (3):

( (-infty, 1) cup (3, infty) ) cap (-infty, 1) = (-infty, 1)

- Next, intersect with (1):

( (-infty, -3] cup [0, infty) ) cap (-infty, 1) = (-infty, -3] cup [0, 1)

Comparing this with

(-infty, -alpha] cup [beta, gamma)

: -

alpha = 3

beta = 0

gamma = 1

Now calculate the target sum:

2alpha + beta + gamma = 2(3) + 0 + 1 = 7

### Pattern Recognition Location of roots: When both roots are positive, checking sum and product signs along with

D ge 0

is the standard and fastest set of inequalities, avoiding complex vertex projections. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Practical Chemistry

Q jee_main_2025_02_april_morning Geometry of Complex Numbers

Let

z

be a complex number such that

|z| = 1

. If

frac2 + k^2zk + overlinez = kz, k in mathbbR

, then the maximum distance of

k + ik^2

from the circle

|z - (1 + 2i)| = 1

is:

A. $sqrt5 + 1$ $\sqrt{5} + 1$
B. $2$ $2$
C. $3$ $3$
D. $sqrt3 + 1$ $\sqrt{3} + 1$

Solution

### Related Formula For a complex number lying on the unit circle:

|z| = 1 implies zoverlinez = 1 implies overlinez = frac1z

Maximum distance from a point

P

to a circle with center

C

and radius

r

is:

d_max = PC + r

### Core Logic Simplify the algebraic condition using

overlinez = 1/z

to uniquely determine the value of the parameter

k

, then compute geometric distances. ### Step 1: Solve for k Cross-multiply the given expression:

2 + k^2z = kz(k + overlinez) = k^2z + kzoverlinez

Since

zoverlinez = |z|^2 = 1

2 + k^2z = k^2z + k(1) implies k = 2

### Step 2: Locate Point and Circle Parameters Substitute

k=2

into the target point expression

P = k + ik^2

P = 2 + 4i equiv (2,4)

The circle equation is

|z - (1 + 2i)| = 1

, which represents a circle centered at

C = (1, 2)

with radius

r = 1

. ### Step 3: Compute Geometric Distances Find the Euclidean distance between

P(2,4)

and center

C(1,2)

PC = sqrt(2-1)^2 + (4-2)^2 = sqrt1 + 4 = sqrt5

The maximum distance from the point to the circle boundary is:

d_max = PC + r = sqrt5 + 1

### Pattern Recognition Notice how

k^2z

cancels perfectly on both sides during expansion due to the unique property of uni-modular complex numbers (

zoverlinez=1

), rendering the calculation of

k

trivial. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Complex Numbers and Quadratic Equations

Q jee_main_2025_02_april_morning Theory of Equations

Let

P_n = alpha^n + beta^n

n in mathbbN

. If

P_10 = 123

P_9 = 76

P_8 = 47

and

P_1 = 1

, then the quadratic equation having roots

frac1alpha

and

frac1beta

is:

A. $x^2 - x + 1 = 0$ $x^2 - x + 1 = 0$
B. $x^2 + x - 1 = 0$ $x^2 + x - 1 = 0$
C. $x^2 - x - 1 = 0$ $x^2 - x - 1 = 0$
D. $x^2 + x + 1 = 0$ $x^2 + x + 1 = 0$

Solution

### Related Formula Newton's Sums for the roots of a quadratic equation

ax^2 + bx + c = 0

a P_n + b P_n-1 + c P_n-2 = 0

### Core Logic Observe the recurrence relation from the given numerical values of

P_n

to construct the base quadratic equation satisfied by

alpha

and

beta

, then invert the roots. ### Step 1: Identify the Linear Recurrence Relation Compare the provided sequence values:

P_8 + P_9 = 47 + 76 = 123 = P_10

This fits the general sequence relation:

P_n = P_n-1 + P_n-2 implies P_n - P_n-1 - P_n-2 = 0

### Step 2: Construct the Base Quadratic Equation The characteristic equation corresponding to this recurrence relation is:

x^2 - x - 1 = 0

Thus,

alpha

and $\beta$ are roots of

x^2 - x - 1 = 0

, giving sum

alpha+beta = 1

and product

alphabeta = -1

(which matches

P_1 = alpha+beta = 1

). ### Step 3: Construct Equation with Reciprocal Roots To find the equation with roots

frac1alpha

and

frac1beta

, apply the transformations:

textSum of new roots = frac1alpha + frac1beta = fracalpha + betaalphabeta = frac1-1 = -1

textProduct of new roots = frac1alphabeta = frac1-1 = -1

The new quadratic equation is:

x^2 - (textSum)x + (textProduct) = 0 implies x^2 - (-1)x + (-1) = 0 implies x^2 + x - 1 = 0

### Pattern Recognition The recurrence pattern

P_n = P_n-1 + P_n-2

is the Fibonacci sequence recurrence line. Its roots generate the Golden Ratio layout from

x^2-x-1=0

. Inverting roots swaps the coefficients of

x^2

and the constant term, yielding

x^2+x-1=0

instantly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Complex Numbers and Quadratic Equations Class 11 Mathematics: Sequences and Series

Q65 jee_main_2025_03_april_evening Geometry of Complex Numbers

z_1, z_2, z_3 in mathbbC

are the vertices of an equilateral triangle, whose centroid is

z_0

, then

sum_k=1^3 (z_k - z_0)^2

is equal to

A. $0$ $0$
B. $1$ $1$
C. $i$ $i$
D. $-i$ $-i$

Solution

### Related Formula For any equilateral triangle with vertices

z_1, z_2, z_3

z_1^2 + z_2^2 + z_3^2 = z_1 z_2 + z_2 z_3 + z_3 z_1 quad text--- (1)

Centroid equation:

z_0 = fracz_1 + z_2 + z_33 implies z_1 + z_2 + z_3 = 3z_0

### Core Logic Let's expand the target sum:

sum_k=1^3 (z_k - z_0)^2 = (z_1 - z_0)^2 + (z_2 - z_0)^2 + (z_3 - z_0)^2

### Step 1: Expansion and Algebraic Grouping Expanding each quadratic term:

= (z_1^2 + z_2^2 + z_3^2) - 2z_0(z_1 + z_2 + z_3) + 3z_0^2

Substitute

z_1 + z_2 + z_3 = 3z_0

= (z_1^2 + z_2^2 + z_3^2) - 2z_0(3z_0) + 3z_0^2

= z_1^2 + z_2^2 + z_3^2 - 3z_0^2

### Step 2: Resolving using Equilateral Condition Substitute

3z_0^2 = 3 left(fracz_1+z_2+z_33right)^2 = frac(z_1+z_2+z_3)^23

= (z_1^2 + z_2^2 + z_3^2) - fracz_1^2 + z_2^2 + z_3^2 + 2(z_1z_2 + z_2z_3 + z_3z_1)3

= frac2(z_1^2 + z_2^2 + z_3^2) - 2(z_1z_2 + z_2z_3 + z_3z_1)3

= frac23(z_1^2 + z_2^2 + z_3^2 - (z_1 z_2 + z_2 z_3 + z_3 z_1))

Using condition (1) for equilateral triangles, the terms inside the parentheses equal

0

. Thus:

= 0

### Pattern Recognition This is a standard invariant of equilateral triangles. Any translation to center of mass coordinates leaves the shape invariant, making the sum of squares of coordinate vectors relative to the centroid equal to zero. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Complex Numbers

Q61 jee_main_2025_07_april_morning Geometry of Complex Numbers

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.

A. $textboth are incorrect$ $\text{both are incorrect}$
B. $textonly (S1) is correct$ $\text{only (S1) is correct}$
C. $textonly (S2) is correct$ $\text{only (S2) is correct}$
D. $textboth are correct$ $\text{both are correct}$

Solution

### 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

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