Let A, B, C be three sets of complex...

MathematicsGeometrical Applications of Complex NumbersJEE Advanced 2008Difficult

Comprehension Passage

Let

A, B, C

be three sets of complex numbers as defined below

A = {z : I m z \geq 1}

B = {z : ∣ z - 2 - i ∣ = 3}

C = {z : R e ((1 - i) z) = 2}

Question 1:

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

(A)

(B)

(C)

(D)

\infty

Answer:B

Question 2:

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

(A)

25 and 29

(B)

30 and 34

(C)

35 and 39

(D)

40 and 44

Answer:C

Question 3:

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

(A)

-6 and 3

(B)

-3 and 6

(C)

-6 and 6

(D)

3 and 9

Answer:D

Visualized Solution (Hindi)

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

Set $A = {z : I m (z) \geq 1}$
Let $z = x + i y$ , then $I m (z) = y$
The condition becomes $y \geq 1$
This represents the half-plane above and including the line $y = 1$ .

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

Set $B = {z : ∣ z - (2 + i) ∣ = 3}$
This is the standard form of a circle $∣ z - z_{0} ∣ = r$
Center: $z_{0} = 2 + i ⟹ (2, 1)$
Radius: $r = 3$
Cartesian equation: $(x - 2)^{2} + (y - 1)^{2} = 9$

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

Set $C = {z : R e ((1 - i) z) = 2}$
$(1 - i) (x + i y) = x + i y - i x - i^{2} y = (x + y) + i (y - x)$
Taking the real part: $x + y = 2$
This is a straight line with intercepts $(2, 0)$ and $(0, 2)$ .

Finding the Intersection .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A \cap B \cap C

To find $B \cap C$ , substitute $y = 2 - x$ into $(x - 2)^{2} + (y - 1)^{2} = 9$
$(x - 2)^{2} + (2 - x - 1)^{2} = 9$
Solving this quadratic gives two values of $x$ .
Constraint from $A$ : $y \geq 1 ⟹ 2 - x \geq 1 ⟹ x \leq 2 - 1$
Only one root satisfies this condition. Thus, $n (A \cap B \cap C) = 1$ .

Question 2: Geometric Interpretation

Evaluate $∣ z - (- 1 + i) ∣^{2} + ∣ z - (5 + i) ∣^{2}$ for $z \in A \cap B \cap C$
Let $Q = - 1 + i ⟹ (- 1, 1)$ and $R = 5 + i ⟹ (5, 1)$
The expression is $P Q^{2} + P R^{2}$
Midpoint of $QR = (\frac{- 1 + 5}{2}, \frac{1 + 1}{2}) = (2, 1)$ , which is the center of circle B.
Thus, $QR$ is a diameter of length $6$ .

Question 2: Applying Pythagoras Theorem

Since $QR$ is a diameter, $∠ QP R = 9 0^{\circ}$ (Angle in a semi-circle).
By Pythagoras Theorem: $P Q^{2} + P R^{2} = Q R^{2}$
$QR = ∣5 - (- 1) ∣ = 6$
Value $= 6^{2} = 36$
$36$ lies between $35$ and $39$ .

Question 3: Bounds for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∣ z ∣ - ∣ w ∣ + 3

$z$ is a fixed point $P$ on circle $B$ .
$w$ satisfies $∣ w - 2 i ∣ < 3$ .
Using Triangle Inequality: $∣∣ z ∣ - ∣ w ∣∣ \leq ∣ z - w ∣$
The maximum possible value of $∣ z - w ∣$ is approximately $6$ based on the diameters.
Thus, $- 6 < ∣ z ∣ - ∣ w ∣ < 6$ .
Adding $3$ : $- 3 < ∣ z ∣ - ∣ w ∣ + 3 < 9$ .
The value lies between $3$ and $9$ .

00:00 / 00:00

Conceptually Similar Problems

MathematicsGeometrical Applications of Complex NumbersJEE Advanced 2013Moderate

View in:English Hindi

Comprehension Passage

Let

S = S_{1} \cap S_{2} \cap S_{3}

, where

S_{1} = {z \in C : ∣ z ∣ < 4}

S_{2} = {z \in C : I m [\frac{z - 1 + 3 i}{1 - 3 i}] > 0}

and

S_{3} = {z \in C : R ez > 0}

Question 1:

Area of $S =$

(A)

f r a c 10 p i 3

(B)

f r a c 20 p i 3

(C)

f r a c 16 p i 3

(D)

f r a c 32 p i 3

Answer:B

Question 2:

$mi n_{z in S} ∣1 - 3 i - z ∣ =$

(A)

f r a c 2 - s q r t 3 2

(B)

f r a c 2 + s q r t 3 2

(C)

f r a c 3 - s q r t 3 2

(D)

f r a c 3 + s q r t 3 2

Answer:C

MathematicsGeometrical Applications of Complex NumbersJEE Advanced 2010Moderate

View in:English Hindi

Match the statements in Column I with those in Column II. [Note : Here $z$ takes values in the complex plane and $I m z$ and $R ez$ denote , respectively, the imaginary part and the real part of $z$ . ]

List-I

(P)

The set of points

z

satisfying

∣ z - i ∣ z ∣∣ = ∣ z + i ∣ z ∣∣

is contained in or equal to

(Q)

The set of points

z

satisfying

∣ z + 4∣ + ∣ z - 4∣ = 10

is contained in or equal to

(R)

∣ w ∣ = 2

, then the set of points

z = w - \frac{1}{w}

is contained in or equal to

(S)

∣ w ∣ = 1

, then the set of points

z = w + \frac{1}{w}

is contained in or equal to

List-II

(1)

an ellipse with eccentricity

\frac{4}{5}

(2)

the set of points

z

satisfying

I m z = 0

(3)

the set of points

z

satisfying

∣ I m z ∣ \leq 1

(4)

the set of points

z

satisfying

∣ R ez ∣ < 2

(5)

the set of points

z

satisfying

∣ z ∣ \leq 3

Answer:Q → 3, Q → 1, R → NaN, S → NaN

MathematicsConjugate and ModulusJEE Advanced 2006Moderate

View in:English Hindi

If $\frac{w - w ˉ z}{1 - z}$ is purely real where $w = α + i β, β \neq = 0$ and $z \neq = 1$ , then the set of the values of $z$ is

(A)

{z : ∣ z ∣ = 1}

(B)

{z : z = \overset{z}{ˉ}}

(C)

{z : z \neq = 1}

(D)

{z : ∣ z ∣ = 1, z \neq = 1}

Answer:D

MathematicsCube Roots and nth Roots of UnityJEE Advanced 2014Moderate

View in:English Hindi

Let $z_{k} = cos (\frac{2 k π}{10}) + i sin (\frac{2 k π}{10}); k = 1, 2, \dots, 9$ . Match the statements in List-I with those in List-II.

List-I

(P)

For each

z_{k}

there exists a

z_{j}

such that

z_{k} \cdot z_{j} = 1

(Q)

There exists a

k \in {1, 2, \dots, 9}

such that

z_{1} \cdot z = z_{k}

has no solution

z

in the set of complex numbers

(R)

\frac{∣1 - z _{1} ∣∣1 - z _{2} ∣ \dots ∣1 - z _{9} ∣}{10}

equals

(S)

1 - \sum_{k = 1}^{9} cos (\frac{2 k π}{10})

equals

List-II

(1)

True

(2)

False

(3)

(4)

Answer:P → 1, Q → 2, R → 3, S → 4

MathematicsGeometrical Applications of Complex NumbersJEE Main 2010Easy

View in:English Hindi

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

(A)

(B)

(C)

\infty

(D)

Answer:A

MathematicsMonotonicityJEE Advanced 2011Difficult

View in:English Hindi

Match the statements given in Column-I with the intervals/union of intervals given in Column-II.

List-I

(P)

The set

{Re (\frac{2 i z}{1 - z ^{2}}) : z is a complex number, ∣ z ∣ = 1, z \neq = \pm 1}

(Q)

The domain of the function

f (x) = sin^{- 1} (\frac{8 ( 3 ) ^{x - 2}}{1 - 3 ^{2 (x - 1)}})

(R)

f (θ) = 1 - tan θ - 1 tan θ 1 - tan θ 1 tan θ 1

, then the set

{f (θ) : 0 \leq θ < \frac{π}{2}}

(S)

f (x) = x^{3/2} (3 x - 10), x \geq 0

then

f (x)

is increasing in

List-II

(1)

(- \infty, - 1) \cup (1, \infty)

(2)

(- \infty, - 0) \cup (0, \infty)

(3)

[2, \infty)

(4)

(- \infty, - 1] \cup [1, \infty)

(5)

(0, 0] \cup [2, \infty)

Answer:P → 4, Q → 5, R → 3, S → 3

MathematicsEuler's Form and De Moivre's TheoremJEE Advanced 1996Moderate

View in:English Hindi

Find all non-zero complex numbers $Z$ satisfying $\overset{ˉ}{Z} = i Z^{2}$ .

Answer:

i, \frac{\pm 3}{2} - \frac{i}{2}

MathematicsGeometrical Applications of Complex NumbersJEE Main 2014Easy

View in:English Hindi

If $z$ is a complex number such that $∣ z ∣ \geq 2$ , then the minimum value of $∣ z + 1/2∣$

(A)

is strictly greater than 5/2

(B)

is strictly greater than 3/2 but less than 5/2

(C)

is equal to 5/2

(D)

lie in the interval (1, 2)

Answer:B

MathematicsConjugate and ModulusJEE Advanced 2000SEasy

View in:English Hindi

If $z_{1}, z_{2}$ and $z_{3}$ are complex numbers such that $∣ z_{1} ∣ = ∣ z_{2} ∣ = ∣ z_{3} ∣ = ∣1/ z_{1} + 1/ z_{2} + 1/ z_{3} ∣ = 1$ , then $∣ z_{1} + z_{2} + z_{3} ∣$ is

(A)

equal to 1

(B)

less than 1

(C)

greater than 3

(D)

equal to 3

Answer:A

MathematicsEuler's Form and De Moivre's TheoremJEE Advanced 1982Moderate

View in:English Hindi

If $z = (\frac{3}{2} + \frac{i}{2})^{5} + (\frac{3}{2} - \frac{i}{2})^{5}$ , then

(A)

Re(z) = 0

(B)

Im(z) = 0

(C)

Re(z) > 0, Im(z) > 0

(D)

Re(z) > 0, Im(z) < 0

Answer:B

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

Visualized Solution (Hindi)

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

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

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

Finding the Intersection .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A∩B∩C

Question 2: Geometric Interpretation

Question 2: Applying Pythagoras Theorem

Question 3: Bounds for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∣z∣−∣w∣+3

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Comprehension Passage

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List-I

List-II

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List-I

List-II

.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

Match the statements given in Column-I with the intervals/union of intervals given in Column-II.

List-I

List-II

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List-I

List-II

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If 1−zw−wˉz​ is purely real where w=α+iβ,β=0 and z=1, then the set of the values of z is

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List-I

List-II

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List-I

List-II

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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

Defining the Region $A$

Visualizing the Circle $B$

Deriving the Line $C$

Finding the Intersection $A \cap B \cap C$

Question 3: Bounds for $∣ z ∣ - ∣ w ∣ + 3$

Area of $S =$

$mi n_{z in S} ∣1 - 3 i - z ∣ =$

If $\frac{w - w ˉ z}{1 - z}$ is purely real where $w = α + i β, β \neq = 0$ and $z \neq = 1$ , then the set of the values of $z$ is

Let $z_{k} = cos (\frac{2 k π}{10}) + i sin (\frac{2 k π}{10}); k = 1, 2, \dots, 9$ . Match the statements in List-I with those in List-II.

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

Find all non-zero complex numbers $Z$ satisfying $\overset{ˉ}{Z} = i Z^{2}$ .

If $z$ is a complex number such that $∣ z ∣ \geq 2$ , then the minimum value of $∣ z + 1/2∣$

If $z_{1}, z_{2}$ and $z_{3}$ are complex numbers such that $∣ z_{1} ∣ = ∣ z_{2} ∣ = ∣ z_{3} ∣ = ∣1/ z_{1} + 1/ z_{2} + 1/ z_{3} ∣ = 1$ , then $∣ z_{1} + z_{2} + z_{3} ∣$ is

If $z = (\frac{3}{2} + \frac{i}{2})^{5} + (\frac{3}{2} - \frac{i}{2})^{5}$ , then

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

Area of $S =$

$mi n_{z in S} ∣1 - 3 i - z ∣ =$

If $\frac{w - w ˉ z}{1 - z}$ is purely real where $w = α + i β, β \neq = 0$ and $z \neq = 1$ , then the set of the values of $z$ is

Let $z_{k} = cos (\frac{2 k π}{10}) + i sin (\frac{2 k π}{10}); k = 1, 2, \dots, 9$ . Match the statements in List-I with those in List-II.

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

Find all non-zero complex numbers $Z$ satisfying $\overset{ˉ}{Z} = i Z^{2}$ .

If $z$ is a complex number such that $∣ z ∣ \geq 2$ , then the minimum value of $∣ z + 1/2∣$

If $z_{1}, z_{2}$ and $z_{3}$ are complex numbers such that $∣ z_{1} ∣ = ∣ z_{2} ∣ = ∣ z_{3} ∣ = ∣1/ z_{1} + 1/ z_{2} + 1/ z_{3} ∣ = 1$ , then $∣ z_{1} + z_{2} + z_{3} ∣$ is

If $z = (\frac{3}{2} + \frac{i}{2})^{5} + (\frac{3}{2} - \frac{i}{2})^{5}$ , then