JEE Advanced Maths – Complex Numbers & Quadratic Equations PYQ with Detailed Solutions

Q 21 :

$Let z_{k} = \cos (\frac{2 k π}{10}) + i \sin (\frac{2 k π}{10}), k = 1, 2, \dots, 9 .$ [2014]

List - I List - II

P. For each $z_{k}$ there exists $z_{j}$ such that $z_{k} . z_{j} = 1$ 1. True

Q. There exists a $k \in {1, 2, \dots, 9}$ such that $z_{1} . z = z_{k}$ has no solution $z$ in the set of complex numbers 2. False

R. $\frac{| 1 - z_{1} | | 1 - z_{2} | \dots | 1 - z_{9} |}{10}$ equals 3. 1

S. $1 - \sum_{k = 1}^{9} \cos (\frac{2 k π}{10})$ equals 4. 2

$P \to 1, Q \to 2, R \to 4, S \to 3$
$P \to 2, Q \to 1, R \to 3, S \to 4$
$P \to 1, Q \to 2, R \to 3, S \to 4$
$P \to 2, Q \to 1, R \to 4, S \to 3$

1

2

3

4

(3)

$(P) \to (1) : z_{k} = \cos \frac{2 k π}{10} + i \sin \frac{2 k π}{10}, k = 1 to 9$

$∴$ $z_{k} = e^{i \frac{2 k π}{10}}$

$Now z_{k} . z_{j} = 1 \Rightarrow z_{j} = \frac{1}{z_{k}} = e^{- i \frac{2 k π}{10}} = \bar{z_{k}}$

$We know if z_{k} is 10th root of unity so will be \bar{z_{k}}$

$∴$ $For every z_{k} there exist z_{i} = \bar{z_{k}}$

$Such that z_{k} . z_{j} = z_{k} . \bar{z_{k}} = 1$

Hence the statement is true.

$(Q) \to (2) : z_{1} = z_{k} \Rightarrow z = \frac{z_{k}}{z_{1}}, z_{1} \neq 0$

$∴$ $We can always find a solution of z_{1} z = z_{k}$

Hence the statement is false.

$(R) \to (3) : We know z^{10} - 1 = (z - 1) (z - z_{1}) \dots (z - z_{9})$

$\Rightarrow (z - z_{1}) (z - z_{2}) \dots (z - z_{9}) = \frac{z^{10} - 1}{z - 1} = 1 + z + z^{2} + \dots + z^{9}$

$For z = 1, we get (1 - z_{1}) (1 - z_{2}) \dots (1 - z_{9}) = 10$

$∴$ $\frac{| 1 - z_{1} | | 1 - z_{2} | \dots | 1 - z_{9} |}{10} = 1$

$(S) \to (4) : 1, z_{1}, z_{2}, \dots, z_{9} are 10th roots of unity.$

$∴$ $z^{10} - 1 = 0$

$From equation 1 + z_{1} + z_{2} + \dots + z_{9} = 0,$

$Re (1) + Re (z_{1}) + Re (z_{2}) + \dots + Re (z_{9}) = 0$

$\Rightarrow Re (z_{1}) + Re (z_{2}) + \dots + Re (z_{9}) = - 1$

$\Rightarrow \sum_{k = 1}^{9} \cos \frac{2 k π}{10} = - 1$ $\Rightarrow 1 - \sum_{k = 1}^{9} \cos \frac{2 k π}{10} = 2$

$Hence (3) is the correct option.$

Q 22 :

Let $A, B, C$ be three sets of complex numbers as defined below

$A = {z : Im (z) \geq 1}$

$B = {z : | z - 2 - i | = 3}$

$C = {z : Re ((1 - i) z) = \sqrt{2}}$

Q. $Let z be any point in A \cap B \cap C .$ $Then,$ $| z + 1 - i |^{2}$ $+$ $| z - 5 - i |^{2}$ $lies between$ [2008]

25 and 29
30 and 34
35 and 39
40 and 44

1

2

3

4

(3)

$Given: A = {z : Im (z) \geq 1} = {(x, y) : y \geq 1}$

$Clearly A is the set of all points lying on or above the line y = 1 in Cartesian plane.$

$B = {z : | z - 2 - i | = 3} = {(x, y) : {(x - 2)}^{2} + {(y - 1)}^{2} = 9}$

$\Rightarrow$ B is the set of all points lying on the boundary of the circle with centre (2, 1) and radius 3.

$C = {z : Re [(1 - i) z] = \sqrt{2}} = {(x, y) : x + y = \sqrt{2}}$

$\Rightarrow$ C is the set of all points lying on the straight line represented by $x + y = \sqrt{2} .$

Graphically, the three sets are represented as shown below:

[IMAGE 12]

$Since, z is a point of A \cap B \cap C \Rightarrow z represents the point P$

$∴$ $| z + 1 - i |^{2} + | z - 5 - i |^{2}$

$\Rightarrow$ $| z - (- 1 + i) |^{2} + | z - (5 + i) |^{2}$

$\Rightarrow P Q^{2} + P R^{2} = Q R^{2} = 6^{2} = 36, which lies between 35 and 39$

$∴$ $(3) is correct option.$

Q 23 :

$Let A, B, C be three sets of complex numbers as defined below$

$A = {z : Im (z) \geq 1}$

$B = {z : | z - 2 - i | = 3}$

$C = {z : Re ((1 - i) z) = \sqrt{2}}$

Q. $Let z be any point A \cap B \cap C and let w be any point satisfying$ $| w - 2 - i |$ $< 3 .$ $Then, | z | - | w | + 3 lies between$ [2008]

- 6 and 3
- 3 and 6
- 6 and 6
- 3 and 9

1

2

3

4

(4)

$Given: A = {z : Im (z) \geq 1} = {(x, y) : y \geq 1}$

$Clearly A is the set of all points lying on or above the line y = 1 in Cartesian plane.$

$B = {z : | z - 2 - i | = 3} = {(x, y) : {(x - 2)}^{2} + {(y - 1)}^{2} = 9}$

$\Rightarrow$ B is the set of all points lying on the boundary of the circle with centre (2, 1) and radius 3.

$C = {z : Re [(1 - i) z] = \sqrt{2}} = {(x, y) : x + y = \sqrt{2}}$

$\Rightarrow$ C is the set of all points lying on the straight line represented by $x + y = \sqrt{2} .$

Graphically, the three sets are represented as shown below:

[IMAGE 13]

$Given$ $| w - 2 - i |$ $< 3$

$\Rightarrow Distance between w and (2 + i), i.e. S is smaller than 3 .$

$\Rightarrow w is a point lying inside the circle with centre S and radius 3 .$

$\Rightarrow Distance between z (i.e. the point P) and w should be smaller than 6 (the diameter of the circle)$

$i.e. | z - w | < 6$

$But we know that$ $| | z | - | w | |$ $\leq | z - w |$

$\Rightarrow$ $| | z | - | w | |$ $< 6 \Rightarrow - 6 < | z | - | w | < 6$

$\Rightarrow - 3 < | z | - | w | + 3 < 9$

Topic Question Set

Q 22 :

Let $A, B, C$ be three sets of complex numbers as defined below

$A = {z : Im (z) \geq 1}$

$B = {z : | z - 2 - i | = 3}$

$C = {z : Re ((1 - i) z) = \sqrt{2}}$

Q. $Let z be any point in A \cap B \cap C .$ $Then,$ $| z + 1 - i |^{2}$ $+$ $| z - 5 - i |^{2}$ $lies between$ [2008]

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	List - I		List - II
P.	For each $z_{k}$ there exists $z_{j}$ such that $z_{k} . z_{j} = 1$	1.	True
Q.	There exists a $k \in {1, 2, \dots, 9}$ such that $z_{1} . z = z_{k}$ has no solution $z$ in the set of complex numbers	2.	False
R.	$\frac{\| 1 - z_{1} \| \| 1 - z_{2} \| \dots \| 1 - z_{9} \|}{10}$ equals	3.	1
S.	$1 - \sum_{k = 1}^{9} \cos (\frac{2 k π}{10})$ equals	4.	2

Topic Question Set

Q 22 : Let A,B,C be three sets of complex numbers as defined below A={z:Im(z)≥1} B={z:|z-2-i|=3} C={z:Re((1-i)z)=2} Q. Let z be any point in A∩B∩C. Then, |z+1-i|2+|z-5-i|2 lies between [2008]

Q 23 : Let A,B,C be three sets of complex numbers as defined below A={z:Im(z)≥1} B={z:|z-2-i|=3} C={z:Re((1-i)z)=2} Q. Let z be any point A∩B∩C and let w be any point satisfying |w-2-i|<3. Then, |z|-|w|+3 lies between [2008]

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Q 22 :

Let $A, B, C$ be three sets of complex numbers as defined below

$A = {z : Im (z) \geq 1}$

$B = {z : | z - 2 - i | = 3}$

$C = {z : Re ((1 - i) z) = \sqrt{2}}$

Q. $Let z be any point in A \cap B \cap C .$ $Then,$ $| z + 1 - i |^{2}$ $+$ $| z - 5 - i |^{2}$ $lies between$ [2008]