Match the statements in Column I with...

MathematicsGeometrical Applications of Complex NumbersJEE Advanced 2010Moderate

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

Visualized Solution (Hindi)

Analyzing Column I(A): .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∣ z - i ∣ z ∣∣ = ∣ z + i ∣ z ∣∣

Given: $∣ z - i ∣ z ∣∣ = ∣ z + i ∣ z ∣∣$
This represents the locus of points $z$ equidistant from $i ∣ z ∣$ and $- i ∣ z ∣$ .
Since $i ∣ z ∣$ and $- i ∣ z ∣$ lie on the imaginary axis, their perpendicular bisector is the Real Axis.
Therefore, $I m (z) = 0$ .
This implies $∣ I m (z) ∣ = 0 \leq 1$ .
Matches: (q) and (r).

Analyzing Column I(B): .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∣ z + 4∣ + ∣ z - 4∣ = 10

Given: $∣ z + 4∣ + ∣ z - 4∣ = 10$
This is the locus of an ellipse where the sum of distances from foci $(\pm 4, 0)$ is $2 a = 10$ .
Major axis: $a = 5$ .
Distance from center to focus: $c = 4$ .
Eccentricity: $e = \frac{c}{a} = \frac{4}{5}$ .
Matches: (p).

Analyzing Column I(C): .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∣ w ∣ = 2, z = w - \frac{1}{w}

Given: $∣ w ∣ = 2$ and $z = w - \frac{1}{w}$ .
Let $w = 2 e^{i θ} = 2 (cos θ + i sin θ)$ .
Then $z = 2 e^{i θ} - \frac{1}{2} e^{- i θ} = (2 - \frac{1}{2}) cos θ + i (2 + \frac{1}{2}) sin θ$ .
$z = \frac{3}{2} cos θ + i \frac{5}{2} sin θ$ .
Locus is the ellipse: $\frac{x ^{2}}{( 3/2 ) ^{2}} + \frac{y ^{2}}{( 5/2 ) ^{2}} = 1$ .

Checking Matches for Column I(C)

For the ellipse $\frac{x ^{2}}{( 3/2 ) ^{2}} + \frac{y ^{2}}{( 5/2 ) ^{2}} = 1$ :
Eccentricity $e = 1 - \frac{( 3/2 ) ^{2}}{( 5/2 ) ^{2}} = 1 - \frac{9}{25} = \frac{4}{5}$ . Matches (p).
$∣ R e (z) ∣ = ∣ \frac{3}{2} cos θ ∣ \leq 1.5 < 2$ . Matches (s).
$∣ z ∣ = \frac{9}{4} cos^{2} θ + \frac{25}{4} sin^{2} θ \leq \frac{5}{2} = 2.5 < 3$ . Matches (t).

Analyzing Column I(D): .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∣ w ∣ = 1, z = w + \frac{1}{w}

Given: $∣ w ∣ = 1$ and $z = w + \frac{1}{w}$ .
Let $w = e^{i θ}$ . Then $z = e^{i θ} + e^{- i θ} = 2 cos θ$ .
Since $z$ is purely real, $I m (z) = 0$ . Matches (q) and (r).
$∣ R e (z) ∣ = ∣2 cos θ ∣ \leq 2$ . Matches (s) and (t).
Note: Even though $∣ R e (z) ∣$ can be $2$ , the set is contained in the given bounds.

Final Summary of Matches

Final Results:
(A) $\to$ (q, r)
(B) $\to$ (p)
(C) $\to$ (p, s, t)
(D) $\to$ (q, r, s, t)

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Conceptually Similar Problems

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Match the conics in Column I with the statements/expressions in Column II.

List-I

(P)

Circle

(Q)

Parabola

(R)

Ellipse

(S)

Hyperbola

List-II

(1)

The locus of the point

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touches the circle

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(2)

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(3)

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(4)

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(5)

Points

z

in the complex plane satisfying

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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)

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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)

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(B)

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(C)

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(D)

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Answer:D

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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}

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

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z

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(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)

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If $ω = \frac{z}{z - \frac{1}{3} i}$ and $∣ ω ∣ = 1$ , then $z$ lies on

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(D)

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(B)

on a circle with centre at the origin

(C)

either on the real axis or on a circle not passing through the origin.

(D)

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The locus of $z$ which lies in shaded region (excluding the boundaries) is best represented by

(A)

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and

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(B)

z : ∣ z - 1∣ > 2

and

∣ a r g (z - 1) ∣ < π /4

(C)

z : ∣ z + 1∣ < 2

and

∣ a r g (z + 1) ∣ < π /2

(D)

z : ∣ z - 1∣ < 2

and

∣ a r g (z + 1) ∣ < π /2

Answer:A

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* Multiple Correct Options

(A)

the circle with radius

1/2 a

and centre

(1/2 a, 0)

for

a > 0, b \neq = 0

(B)

the circle with radius

- 1/2 a

and centre

(- 1/2 a, 0)

for

a < 0, b \neq = 0

(C)

the x-axis for

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(D)

the y-axis for

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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)

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The complex numbers $z = x + i y$ which satisfy the equation $\frac{z - 5 i}{z + 5 i} = 1$ lie on

(A)

the x-axis

(B)

the straight line

y = 5

(C)

a circle passing through the origin

(D)

none of these

Answer:A

List-I

List-II

Visualized Solution (Hindi)

Analyzing Column I(A): .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∣z−i∣z∣∣=∣z+i∣z∣∣

Analyzing Column I(B): .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∣z+4∣+∣z−4∣=10

Analyzing Column I(C): .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∣w∣=2,z=w−w1​

Checking Matches for Column I(C)

Analyzing Column I(D): .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∣w∣=1,z=w+w1​

Final Summary of Matches

Conceptually Similar Problems

Match the conics in Column I with the statements/expressions in Column II.

List-I

List-II

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; } If ∣z2−1∣=∣z∣2+1, then z lies on

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let zk​=cos(102kπ​)+isin(102kπ​);k=1,2,…,9. Match the statements in List-I with those in List-II.

List-I

List-II

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If ω=z−31​iz​ and ∣ω∣=1, then z lies on

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If z=1 and z−1z2​ is real, then the point represented by the complex number z lies

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The locus of z which lies in shaded region (excluding the boundaries) is best represented by

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let a,b∈R and a2+b2=0. Suppose S={z∈C:z=a+ibt1​,t∈R,t=0}, where i=−1​. If z=x+iy and z∈S, then (x,y) lies on

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

List-I

List-II

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The complex numbers z=x+iy which satisfy the equation ​z+5iz−5i​​=1 lie on

List-I

List-II

Match the conics in Column I with the statements/expressions in Column II.

List-I

List-II

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; } If ∣z2−1∣=∣z∣2+1, then z lies on

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let zk​=cos(102kπ​)+isin(102kπ​);k=1,2,…,9. Match the statements in List-I with those in List-II.

List-I

List-II

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If ω=z−31​iz​ and ∣ω∣=1, then z lies on

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If z=1 and z−1z2​ is real, then the point represented by the complex number z lies

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The locus of z which lies in shaded region (excluding the boundaries) is best represented by

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let a,b∈R and a2+b2=0. Suppose S={z∈C:z=a+ibt1​,t∈R,t=0}, where i=−1​. If z=x+iy and z∈S, then (x,y) lies on

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

List-I

List-II

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The complex numbers z=x+iy which satisfy the equation ​z+5iz−5i​​=1 lie on

Analyzing Column I(A): $∣ z - i ∣ z ∣∣ = ∣ z + i ∣ z ∣∣$

Analyzing Column I(B): $∣ z + 4∣ + ∣ z - 4∣ = 10$

Analyzing Column I(C): $∣ w ∣ = 2, z = w - \frac{1}{w}$

Analyzing Column I(D): $∣ w ∣ = 1, z = w + \frac{1}{w}$

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

If $∣ z^{2} - 1∣ = ∣ z ∣^{2} + 1$ , then $z$ lies on

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.

If $ω = \frac{z}{z - \frac{1}{3} i}$ and $∣ ω ∣ = 1$ , then $z$ lies on

If $z \neq = 1$ and $\frac{z ^{2}}{z - 1}$ is real, then the point represented by the complex number $z$ lies

The locus of $z$ which lies in shaded region (excluding the boundaries) is best represented by

Let $a, b \in R$ and $a^{2} + b^{2} \neq = 0$ . Suppose $S = {z \in C : z = \frac{1}{a + ib t}, t \in R, t \neq = 0}$ , where $i = - 1$ . If $z = x + i y$ and $z \in S$ , then $(x, y)$ lies on

The complex numbers $z = x + i y$ which satisfy the equation $\frac{z - 5 i}{z + 5 i} = 1$ lie on

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

If $∣ z^{2} - 1∣ = ∣ z ∣^{2} + 1$ , then $z$ lies on

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.

If $ω = \frac{z}{z - \frac{1}{3} i}$ and $∣ ω ∣ = 1$ , then $z$ lies on

If $z \neq = 1$ and $\frac{z ^{2}}{z - 1}$ is real, then the point represented by the complex number $z$ lies

The locus of $z$ which lies in shaded region (excluding the boundaries) is best represented by

Let $a, b \in R$ and $a^{2} + b^{2} \neq = 0$ . Suppose $S = {z \in C : z = \frac{1}{a + ib t}, t \in R, t \neq = 0}$ , where $i = - 1$ . If $z = x + i y$ and $z \in S$ , then $(x, y)$ lies on

The complex numbers $z = x + i y$ which satisfy the equation $\frac{z - 5 i}{z + 5 i} = 1$ lie on