Let z_k = cos(2kπ/10) + i sin(2kπ/10);...

MathematicsCube Roots and nth Roots of UnityJEE Advanced 2014Moderate

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

Visualized Solution (Hindi)

Visualizing the .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 1 0^{t h} Roots of Unity

Given $z_{k} = cos (\frac{2 k π}{10}) + i sin (\frac{2 k π}{10}) = e^{i \frac{2 k π}{10}}$ for $k = 1, 2, \dots, 9$ .
These are the non-real $1 0^{t h}$ roots of unity.
The complete set of $1 0^{t h}$ roots of unity is ${1, z_{1}, z_{2}, \dots, z_{9}}$ .
Geometrically, these points lie on a unit circle $∣ z ∣ = 1$ .

Statement P: Existence of Inverses

Condition: $z_{k} \cdot z_{j} = 1 ⟹ z_{j} = \frac{1}{z _{k}}$ .
Since $∣ z_{k} ∣ = 1$ , we know that $\frac{1}{z _{k}} = \overset{z}{ˉ}_{k}$ .
The conjugate of $e^{i \frac{2 k π}{10}}$ is $e^{- i \frac{2 k π}{10}}$ , which is equivalent to $e^{i \frac{2 ( 10 - k ) π}{10}} = z_{10 - k}$ .
For any $k \in {1, \dots, 9}$ , the index $j = 10 - k$ is also in the set ${1, \dots, 9}$ .
Thus, the statement is True.

Statement Q: Solutions to .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } z_{1} \cdot z = z_{k}

Equation: $z_{1} \cdot z = z_{k}$ .
Since $z_{1} = e^{i \frac{2 π}{10}} \neq = 0$ , we can always solve for $z$ .
$z = \frac{z _{k}}{z _{1}} = e^{i \frac{2 ( k - 1 ) π}{10}}$ .
This $z$ is always a valid complex number for any $k$ .
Thus, the statement is False.

Statement R: Product of Chord Lengths

The $1 0^{t h}$ roots of unity satisfy $z^{10} - 1 = 0$ .
Factorization: $z^{10} - 1 = (z - 1) (z - z_{1}) (z - z_{2}) \dots (z - z_{9})$ .
Dividing by $(z - 1)$ gives: $\frac{z ^{10} - 1}{z - 1} = (z - z_{1}) (z - z_{2}) \dots (z - z_{9})$ .
Also, by geometric series: $\frac{z ^{10} - 1}{z - 1} = 1 + z + z^{2} + \dots + z^{9}$ .

Evaluating the Product Limit

Substitute $z = 1$ into the identity:
$1 + 1 + 1^{2} + \dots + 1^{9} = (1 - z_{1}) (1 - z_{2}) \dots (1 - z_{9})$ .
$10 = (1 - z_{1}) (1 - z_{2}) \dots (1 - z_{9})$ .
Taking the absolute value: $∣1 - z_{1} ∣∣1 - z_{2} ∣ \dots ∣1 - z_{9} ∣ = 10$ .
The required value is $\frac{10}{10} = 1$ .

Statement S: Sum of Cosines

Property: The sum of all $n^{t h}$ roots of unity is $0$ .
$1 + z_{1} + z_{2} + \dots + z_{9} = 0$ .
Taking the real part: $Re (1 + \sum_{k = 1}^{9} z_{k}) = 0$ .
$1 + \sum_{k = 1}^{9} cos (\frac{2 k π}{10}) = 0 ⟹ \sum_{k = 1}^{9} cos (\frac{2 k π}{10}) = - 1$ .
The expression $1 - \sum_{k = 1}^{9} cos (\frac{2 k π}{10}) = 1 - (- 1) = 2$ .

Final Summary and Takeaway

Final Match:
(P) $\to$ (1): True
(Q) $\to$ (2): False
(R) $\to$ (3): 1
(S) $\to$ (4): 2
Key Takeaway: Use the polynomial $z^{n} - 1 = (z - 1) \prod (z - z_{k})$ for products and $\sum z_{k} = 0$ for sums.

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Conceptually Similar Problems

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Let $ω$ be the complex number $cos \frac{2 π}{3} + i sin \frac{2 π}{3}$ . Then the number of distinct complex numbers $z$ satisfying $z + 1 ω ω^{2} ω z + ω^{2} 1 ω^{2} 1 z + ω = 0$ is equal to

Answer:1

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Prove that there exists no complex number $z$ such that $∣ z ∣ < 1/3$ and $\sum_{r = 1}^{n} a_{r} z^{r} = 1$ where $∣ a_{r} ∣ < 2$ .

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

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(D)

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(B)

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Answer:D

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Match List I with List II:

List-I

(P)

[\frac{1}{y ^{2}} (\frac{c o s ( t a n ^{- 1} y ) + y s i n ( t a n ^{- 1} y )}{c o t ( s i n ^{- 1} y ) + t a n ( s i n ^{- 1} y )})^{2} + y^{4}]^{1/2}

takes value

(Q)

cos x + cos y + cos z = 0 = sin x + sin y + sin z

then possible value of

cos \frac{x - y}{2}

(R)

cos (\frac{π}{4} - x) cos 2 x + sin x sin 2 x sec x = cos x sin 2 x sec x + cos (\frac{π}{4} + x) cos 2 x

then possible value of

sec x

(S)

cot (sin^{- 1} 1 - x^{2}) = sin (tan^{- 1} (x 6))

x \neq = 0

, then possible value of

x

List-II

(1)

\frac{1}{2} \frac{5}{3}

(2)

2

(3)

\frac{1}{2}

(4)

Answer:P → 4, Q → 3, R → 2, S → 1

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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

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Let $z$ and $ω$ be two complex numbers such that $∣ z ∣ \leq 1, ∣ ω ∣ \leq 1$ and $∣ z + iω ∣ = ∣ z - i \overset{ω}{ˉ} ∣ = 2$ then $z$ equals

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1 or i

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(C)

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(D)

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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)

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

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Match the statements/expressions given in Column-I with the values given in Column-II.

List-I

(P)

The number of solutions of the equation

x e^{s i n x} - cos x = 0

in the interval

(0, \frac{π}{2})

(Q)

Value(s) of

k

for which the planes

k x + 4 y + z = 0

4 x + k y + 2 z = 0

and

2 x + 2 y + z = 0

intersect in a straight line

(R)

Value(s) of

k

for which

∣ x - 1∣ + ∣ x - 2∣ + ∣ x + 1∣ + ∣ x + 2∣ = 4 k

has integer solution(s)

(S)

y^{'} = y + 1

and

y (0) = 1

, then value(s) of

y (ln 2)

List-II

(1)

(2)

(3)

(4)

(5)

Answer:P → 1, Q → 4, R → NaN, S → 3

MathematicsGeometrical Applications of Complex NumbersJEE Advanced 1985Moderate

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If $z_{1} = a + ib$ and $z_{2} = c + i d$ are complex numbers such that $∣ z_{1} ∣ = ∣ z_{2} ∣ = 1$ and $R e (z_{1} \overset{z}{ˉ}_{2}) = 0$ , then the pair of complex numbers $w_{1} = a + i c$ and $w_{2} = b + i d$ satisfies

* Multiple Correct Options

(A)

∣ w_{1} ∣ = 1

(B)

∣ w_{2} ∣ = 1

(C)

R e (w_{1} \overset{w}{ˉ}_{2}) = 0

(D)

none of these

Answer:A, B, C

.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

Visualized Solution (Hindi)

Visualizing the .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 10th Roots of Unity

Statement P: Existence of Inverses

Statement Q: Solutions to .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } z1​⋅z=zk​

Statement R: Product of Chord Lengths

Evaluating the Product Limit

Statement S: Sum of Cosines

Final Summary and Takeaway

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List-II

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Statement Q: Solutions to $z_{1} \cdot z = z_{k}$

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