If z 1 , z 2 are the roots of the equation a z 2 + b z + c = 0 , with a , b , c > 0 ; 2 b 2 > 4 a c > b 2 ; z 1 ∈ third quadrant ; z 2 ∈ second quadrant in the argand's plane, then show that arg z 1 z 2 = 2 cos - 1 b 2 4 a c 1 / 2 .

Given, z 1 , z 2 are the roots of the equation a z 2 + b z + c = 0 . ⇒ z 1 , z 2 = - b ± b 2 - 4 a c 2 a ⇒ z 1 , z 2 = - b ± i 4 a c - b 2 2 a ( Given, b 2 - 4 a c < 0 ) Let, z 1 = - b - i 4 a c - b 2 2 a , z 2 = - b + i 4 a c - b 2 2 a & arg z 1 z 2 = θ Now, z 1 z 2 = z 1 z 2 e i θ ⇒ b + i 4 a c - b 2 b - i 4 a c - b 2 = b + i 4 a c - b 2 b - i 4 a c - b 2 e i θ ⇒ b + i 4 a c - b 2 b - i 4 a c - b 2 = e i θ 1 Apply componendo and dividendo ⇒ b + i 4 a c - b 2 + b - i 4 a c - b 2 b + i 4 a c - b 2 - b - i 4 a c - b 2 = e i θ + 1 e i θ - 1 ⇒ 2 b 2 i 4 a c - b 2 = cos θ + 1 + i sin θ cos θ - 1 + i sin θ ⇒ b i 4 a c - b 2 = 2 cos 2 θ 2 - 1 + 1 + i 2 sin θ 2 . cos θ 2 1 - 2 sin 2 θ 2 - 1 + i 2 sin θ 2 . cos θ 2 ⇒ b i 4 a c - b 2 = cos θ 2 i sin θ 2 × cos θ 2 + i sin θ 2 cos θ 2 + i sin θ 2 ⇒ b 4 a c - b 2 2 = cos θ 2 sin θ 2 2 ⇒ b 2 4 a c - b 2 = cos 2 θ 2 sin 2 θ 2 ⇒ 4 a c - b 2 b 2 = tan 2 θ 2 ⇒ 4 a c b 2 - 1 = sec 2 θ 2 - 1 ⇒ 4 a c b 2 = sec 2 θ 2 ⇒ cos 2 θ 2 = b 2 4 a c ⇒ cos θ 2 = b 2 4 a c ⇒ θ = 2 cos - 1 b 2 4 a c ⇒ arg z 1 z 2 = 2 cos - 1 b 2 4 a c

Prove That : a cos x + C 1 n cos 2 x + C 2 n cos 3 x + … . + C n n cos n + 1 x = 2 n · cos n x 2 · cos n + 2 2 x

We know that ( 1 + y ) n = C 0 n + C 1 n y + C 2 n y 2 + … + C n n y n ⇒ y ( 1 + y ) n = C 0 n y + C 1 n y 2 + C 2 n y 3 + … + C n n y n + 1 Put, y = cos x + i sin x ⇒ cos x + i sin x ( 1 + cos x + i sin x ) n = C 0 n cos x + i sin x + C 1 n cos x + i sin x 2 + C 2 n cos x + i sin x 3 + C 3 n cos x + i sin x 4 + … + C n n cos x + i sin x n + 1 ⇒ cos x + i sin x 1 + 2 cos 2 x 2 - 1 + i 2 sin x 2 · cos x 2 n = cos x + i sin x + C 1 n cos 2 x + i sin 2 x + C 2 n cos 3 x + i sin 3 x + C 3 n cos 4 x + i sin 4 x + … + C n n cos n + 1 x + i sin n + 1 x ⇒ 2 n cos n x 2 cos x + i sin x cos x 2 + i sin x 2 n = cos x + C 1 n cos 2 x + C 2 n cos 3 x + C 3 n cos 4 x + … + C n n cos n + 1 x + i sin x + C 1 n sin 2 x + C 2 n sin 3 x + C 3 n sin 4 x + … + C n n sin n + 1 x ⇒ 2 n cos n x 2 cos n + 2 x 2 + i sin n + 2 x 2 = cos x + C 1 n cos 2 x + C 2 n cos 3 x + C 3 n cos 4 x + … + C n n cos n + 1 x + i sin x + C 1 n sin 2 x + C 2 n sin 3 x + C 3 n sin 4 x + … + C n n sin n + 1 x ⇒ cos x + C 1 n cos 2 x + C 2 n cos 3 x + C 3 n cos 4 x + … + C n n cos n + 1 x = 2 n cos n x 2 · cos n + 2 x 2 and sin x + C 1 n sin 2 x + C 2 n sin 3 x + C 3 n sin 4 x + … + C n n sin n + 1 x = 2 n cos n x 2 · sin n + 2 x 2 .

Prove that : b sin x + C 1 n sin 2 x + C 2 n sin 3 x + … … + C n n sin n + 1 x = 2 n · cos n x 2 · sin n + 2 2 x

We know that ( 1 + y ) n = C 0 n + C 1 n y + C 2 n y 2 + … + C n n y n ⇒ y ( 1 + y ) n = C 0 n y + C 1 n y 2 + C 2 n y 3 + … + C n n y n + 1 Put, y = cos x + i sin x ⇒ cos x + i sin x ( 1 + cos x + i sin x ) n = C 0 n cos x + i sin x + C 1 n cos x + i sin x 2 + C 2 n cos x + i sin x 3 + C 3 n cos x + i sin x 4 + … + C n n cos x + i sin x n + 1 ⇒ cos x + i sin x 1 + 2 cos 2 x 2 - 1 + i 2 sin x 2 · cos x 2 n = cos x + i sin x + C 1 n cos 2 x + i sin 2 x + C 2 n cos 3 x + i sin 3 x + C 3 n cos 4 x + i sin 4 x + … + C n n cos n + 1 x + i sin n + 1 x ⇒ 2 n cos n x 2 cos x + i sin x cos x 2 + i sin x 2 n = cos x + C 1 n cos 2 x + C 2 n cos 3 x + C 3 n cos 4 x + … + C n n cos n + 1 x + i sin x + C 1 n sin 2 x + C 2 n sin 3 x + C 3 n sin 4 x + … + C n n sin n + 1 x ⇒ 2 n cos n x 2 cos n + 2 x 2 + i sin n + 2 x 2 = cos x + C 1 n cos 2 x + C 2 n cos 3 x + C 3 n cos 4 x + … + C n n cos n + 1 x + i sin x + C 1 n sin 2 x + C 2 n sin 3 x + C 3 n sin 4 x + … + C n n sin n + 1 x ⇒ cos x + C 1 n cos 2 x + C 2 n cos 3 x + C 3 n cos 4 x + … + C n n cos n + 1 x = 2 n cos n x 2 · cos n + 2 x 2 and sin x + C 1 n sin 2 x + C 2 n sin 3 x + C 3 n sin 4 x + … + C n n sin n + 1 x = 2 n cos n x 2 · sin n + 2 x 2 .

Evaluate: ∑ p = 1 32 3 p + 2 ∑ q = 1 10 sin 2 q π 11 - i cos 2 q π 11 p

We know that cos θ + i sin θ = e i θ Let, L = ∑ p = 1 32 3 p + 2 ∑ q = 1 10 sin 2 q π 11 - i cos 2 q π 11 p ⇒ L = ∑ p = 1 32 3 p + 2 ∑ q = 1 10 - i cos 2 q π 11 + i sin 2 q π 11 p ⇒ L = ∑ p = 1 32 3 p + 2 - i ∑ q = 1 10 e i 2 q π 11 p ⇒ L = ∑ p = 1 32 3 p + 2 - i e i 2 π 11 + e i 4 π 11 + e i 6 π 11 + . . . . + e i 20 π 11 p ⇒ L = ∑ p = 1 32 3 p + 2 - i e i 2 π 11 e i 20 π 11 - 1 e i 2 π 11 - 1 p ⇒ L = ∑ p = 1 32 3 p + 2 - i e i 2 π - e i 2 π 11 e i 2 π 11 - 1 p ⇒ L = ∑ p = 1 32 3 p + 2 - i cos 2 π + i · sin 2 π - e i 2 π 11 e i 2 π 11 - 1 p ⇒ L = ∑ p = 1 32 3 p + 2 - i 1 + 0 - e i 2 π 11 e i 2 π 11 - 1 p ⇒ L = ∑ p = 1 32 3 p + 2 i p ⇒ L = 3 ∑ p = 1 32 p i p + 2 ∑ p = 1 32 i p ⇒ L = 3 i + 2 · i 2 + 3 · i 3 + 4 · i 4 + . . . . + 32 · i 32 + 2 i + i 2 + i 3 + i 4 + . . . . + i 32 ⇒ L = 3 i 1 + 5 + 9 + 13 + . . . . + 29 + - 1 2 + 6 + 10 + . . . + 30 + - i 3 + 7 + 11 + . . . + 31 + 1 · 4 + 8 + 12 + . . . . + 32 + 2 i · i 32 - 1 i - 1 ⇒ L = 3 i 8 2 1 + 29 + - 1 8 2 2 + 30 + - i 8 2 3 + 31 + 8 2 4 + 32 + 2 i · 1 - 1 i - 1 ⇒ L = 3 120 i - 128 - 136 i + 144 + 0 ⇒ L = 48 1 - i

E M B I B E

Beta Question Bank for Engineering: Mathematics>Chapter 6 - Complex Numbers>EXERCISE-4

Embibe Experts Solutions for Chapter: Complex Numbers, Exercise 4: EXERCISE-4

Author:Embibe Experts

Embibe Experts Mathematics Solutions for Exercise - Embibe Experts Solutions for Chapter: Complex Numbers, Exercise 4: EXERCISE-4

Attempt the free practice questions on Chapter 6: Complex Numbers, Exercise 4: EXERCISE-4 with hints and solutions to strengthen your understanding. Beta Question Bank for Engineering: Mathematics solutions are prepared by Experienced Embibe Experts.

Questions from Embibe Experts Solutions for Chapter: Complex Numbers, Exercise 4: EXERCISE-4 with Hints & Solutions

MEDIUM

JEE Main/Advance

IMPORTANT

Beta Question Bank for Engineering: Mathematics>Chapter 6 - Complex Numbers>EXERCISE-4>Q 18

Let $z = x + i y$ be a complex number, where $x$ and $y$ are real numbers. Let A and $B$ be the sets defined by $A = \{z : |z| \leq 2\}$ and $B = \{z : (1 - i) z + (1 + i) \bar{z} \geq 4\}$ Find the area of the region $A \cap B$ .

HARD

JEE Main/Advance

IMPORTANT

Beta Question Bank for Engineering: Mathematics>Chapter 6 - Complex Numbers>EXERCISE-4>Q 19

For all real numbers $x$ , let the mapping $f (x) = \frac{1}{x - i},$ where $i = \sqrt{- 1}$ . If there exist real number $a, b, c$ and $d$ for which $f (a), f (b), f (c)$ and $f (d)$ form a square on the complex plane. Find the area of the square.

HARD

JEE Main/Advance

IMPORTANT

Beta Question Bank for Engineering: Mathematics>Chapter 6 - Complex Numbers>EXERCISE-4>Q 21

If $z_{1}, z_{2}$ are the roots of the equation $a z^{2} + b z + c = 0$ , with $a, b, c > 0; 2 b^{2} > 4 a c > b^{2}; z_{1} \in$ third quadrant ; $z_{2} \in$ second quadrant in the argand's plane, then show that $\arg (\frac{z_{1}}{z_{2}}) = 2 \cos^{- 1} {(\frac{b^{2}}{4 a c})}^{1 / 2}$ .

MEDIUM

JEE Main/Advance

IMPORTANT

Beta Question Bank for Engineering: Mathematics>Chapter 6 - Complex Numbers>EXERCISE-4>Q 23

Prove That : $(a) \cos x + {}^{n}C_{1} \cos 2 x + {}^{n}C_{2} \cos 3 x + \dots . + {}^{n}C_{n} \cos (n + 1) x = 2^{n} \cdot \cos^{n} \frac{x}{2} \cdot \cos (\frac{n + 2}{2}) x$

MEDIUM

JEE Main/Advance

IMPORTANT

Beta Question Bank for Engineering: Mathematics>Chapter 6 - Complex Numbers>EXERCISE-4>Q 24

Prove that : $(b) \sin x + {}^{n}C_{1} \sin 2 x + {}^{n}C_{2} \sin 3 x + \dots \dots + {}^{n}C_{n} \sin (n + 1) x = 2^{n} \cdot \cos^{n} \frac{x}{2} \cdot \sin (\frac{n + 2}{2}) x$

HARD

JEE Main/Advance

IMPORTANT

Beta Question Bank for Engineering: Mathematics>Chapter 6 - Complex Numbers>EXERCISE-4>Q 25

The points $A, B, C$ depict the complex numbers $z_{1}, z_{2}, z_{3}$ respectively on a complex plane & the angle $B & C$ of the triangle $A B C$ are each equal to $\frac{1}{2} (π - α)$ . Show that : ${(z_{2} - z_{3})}^{2} = 4 (z_{3} - z_{1}) (z_{1} - z_{2}) \sin^{2} \frac{α}{2}$

HARD

JEE Main/Advance

IMPORTANT

Beta Question Bank for Engineering: Mathematics>Chapter 6 - Complex Numbers>EXERCISE-4>Q 26

Evaluate: $\sum_{p = 1}^{32} (3 p + 2) {(\sum_{q = 1}^{10} (\sin \frac{2 q π}{11} - i \cos \frac{2 q π}{11}))}^{p}$

MEDIUM

JEE Main/Advance

IMPORTANT

Beta Question Bank for Engineering: Mathematics>Chapter 6 - Complex Numbers>EXERCISE-4>Q 27

Let $P (z) = z^{3} + a z^{2} + b z + c$ , where $a, b$ and $c$ are real. There exist a complex number $w$ such that the three roots of $P (z)$ are $ω + 3 i, ω + 9 i$ and $2 ω - 4$ where $i^{2} = - 1$ then $| a + b + c |$ is equal to

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