Let z 1 = 10 + 6 i and z 2 = 4 + 6 i . If z is any complex number such that the argument of z - z 1 z - z 2 is π 4 , then find the length of the arc of the locus.

Assume, z = x + i y Given, arg z - z 1 z - z 2 = π 4 ⇒ arg x - 10 + i y - 6 x - 4 + i y - 6 = π 4 ⇒ arg x - 10 + i y - 6 x - 4 - i y - 6 x - 4 + i y - 6 x - 4 - i y - 6 = π 4 ⇒ arg x - 10 x - 4 + y - 6 2 + i y - 6 x - 4 - y - 6 x - 10 x - 4 2 + y - 6 2 = π 4 ⇒ y - 6 x - 4 - y - 6 x - 10 x - 4 x - 10 + y - 6 2 = 1 ⇒ x y - 6 x - 4 y + 24 - x y + 6 x + 10 y - 60 = x 2 - 14 x + 40 + y 2 - 12 y + 36 ⇒ x 2 + y 2 - 18 y - 14 x + 112 = 0 ⇒ x - 7 2 + y - 9 2 = 18 So, the radius of the circle is 18 . Hence, the length of the arc or the perimeter is 2 π 18 = 6 π 2 .

Let z 1 = 10 + 6 i and z 2 = 4 + 6 i . If z is any complex number such that the argument of z - z 1 z - z 2 is π 4 , then find the length of the arc of the locus.

Assume, z = x + i y Given, arg z - z 1 z - z 2 = π 4 ⇒ arg x - 10 + i y - 6 x - 4 + i y - 6 = π 4 ⇒ arg x - 10 + i y - 6 x - 4 - i y - 6 x - 4 + i y - 6 x - 4 - i y - 6 = π 4 ⇒ arg x - 10 x - 4 + y - 6 2 + i y - 6 x - 4 - y - 6 x - 10 x - 4 2 + y - 6 2 = π 4 ⇒ y - 6 x - 4 - y - 6 x - 10 x - 4 x - 10 + y - 6 2 = 1 ⇒ x y - 6 x - 4 y + 24 - x y + 6 x + 10 y - 60 = x 2 - 14 x + 40 + y 2 - 12 y + 36 ⇒ x 2 + y 2 - 18 y - 14 x + 112 = 0 ⇒ x - 7 2 + y - 9 2 = 18 So, the radius of the circle is 18 . Hence, the length of the arc or the perimeter is 2 π 18 = 6 π 2 .

Show that z z ¯ + ( 4 - 3 i ) z + ( 4 + 3 i ) z ¯ + 5 = 0 represents circle. Hence find centre and radius.

Given, z z ¯ + ( 4 - 3 i ) z + ( 4 + 3 i ) z ¯ + 5 = 0 . Let z = x + i y So, ( x + i y ) ( x − i y ) + ( 4 − 3 i ) ( x + i y ) + ( 4 + 3 i ) ( x − i y ) + 5 = 0 ⇒ x 2 + y 2 + 4 x + 4 i y − 3 i x + 3 y + 4 x − 4 i y + 3 i x + 3 y + 5 = 0 ⇒ x 2 + y 2 + 8 x + 6 y + 5 = 0 ⇒ x + 4 2 + y + 3 2 = 20 So, the centre of the circle is - 4 , - 3 and the radius of the circle is 20 = 2 5 .

If z 1 & z 2 are two complex numbers & if arg z 1 + z 2 z 1 - z 2 = π 2 but z 1 + z 2 ≠ z 1 - z 2 then identify the figure formed by the points represented by 0 , z 1 , z 2 & z 1 + z 2 .

z 1 and z 2 are two complex number which can be represented by vectors z 1 → and z 2 → in argand plane. Then z 1 → + z 2 → and z 1 → - z 2 → vectors can be represented by diagonals of a parallelogram having z 1 → and z 2 → as its adjacent sides. Now, given arg z 1 + z 2 z 1 - z 2 = π 2 or arg z 1 + z 2 - arg z 1 - z 2 = π 2 Angle between diagonals of parallelogram is π 2 . Also given z 1 + z 2 ≠ z 1 - z 2 . This shows that length of diagonals are not equal. So, the quadrilateral formed is a rhombus but not a square.

Show that 1 - ω + ω 2 1 - ω 2 + ω 4 1 - ω 4 + ω 8 … … … . to 2 n factors = 2 2 n , where ω is the cube root of unity.

We know, 1 - ω + ω 2 = - 2 ω + 1 + ω + ω 2 = - 2 ω 1 - ω 2 + ω 4 = 1 + ω + ω 2 - 2 ω 2 = - 2 ω 2 Similarly, 1 - ω 4 + ω 8 = - 2 ω So, 1 - ω + ω 2 1 - ω 2 + ω 4 1 - ω 4 + ω 8 … … … . 2 n factors = - 2 ω × - 2 ω 2 × - 2 ω × . . . . = 4 ω 3 n = 4 n

Let ω is non-real root of x 3 = 1 . (i) If P = ω n , ( n ∈ N ) and Q = C 0 2 n + C 3 2 n + … … . . + ( 2 n C 1 + C 4 2 n + … … … ω + C 2 2 n + C 5 2 n + . . . . ω 2 , then find P Q . (ii) If P = 1 - ω 2 + ω 2 4 - ω 3 8 … … upto ∞ terms and Q = 1 - ω 2 2 , then find value of P Q .

( i ) Given, Q = C 0 2 n + C 3 2 n + … … . . + ( 2 n C 1 + C 4 2 n + … … … ω + C 2 2 n + C 5 2 n + . . . . ω 2 Put n = 1 , Q = C 0 2 + 2 C 1 ω + 2 C 2 ω 2 = 1 + 2 ω + ω 2 = ω Put n = 2 Q = C 0 4 + C 3 4 + 4 C 1 + C 4 4 ω + C 2 4 ω 2 = 5 + 5 ω + 6 ω 2 = ω 2 Put n = 3 Q = C 0 6 + C 3 6 + C 6 6 + ( 6 C 1 + C 4 6 + … … … ω + 6 C 2 + C 5 6 + … … . . ω 2 ) = 22 + 21 ω + 21 ω 2 = ω 3 Similarly, we can say that Q = ω n So, P Q = 1 . ( ii ) P = 1 - ω 2 + ω 2 4 - ω 3 8 … … ⇒ P = 1 1 + ω 2 = 2 2 + ω = 2 1 - ω 2 So, P Q = 2 1 - ω 2 × 1 - ω 2 2 = 1 .

If x = 1 + i 3 ; y = 1 - i 3 and z = 2 , then prove that x p + y ρ = z ρ for every prime p > 3 .

x = 1 + i 3 = 2 1 2 + i 3 2 = 2 cos θ + i sin θ , where θ = π 3 & y = 1 − i 3 = 2 1 2 − i 3 2 = 2 cos − θ + i sin - θ , where θ = π 3 . Thus using de Moivre's theorem, x P = 2 P ( cos P θ + i sin P θ ) & y P = 2 P ( cos ( − P θ ) + i sin ( − P θ ) ) ∴ x P + y P = 2 P ( 2 cos P θ ) = 2 P 2 × cos P π 3 = 2 P 2 × 1 2 = z P

E M B I B E

Alpha Question Bank for Engineering: Mathematics>Chapter 6 - Complex Numbers>Exercise-1>Q 31

HARD

JEE Main/Advance

IMPORTANT

Earn 100

Let $z_{1} = 10 + 6 i$ and $z_{2} = 4 + 6 i$ . If $z$ is any complex number such that the argument of $\frac{(z - z_{1})}{(z - z_{2})}$ is $\frac{π}{4}$ , then find the length of the arc of the locus.

Important Questions on Complex Numbers

HARD

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 6 - Complex Numbers>Exercise-1>Q 32

Let

I :

Arg (\frac{z - 8 i}{z + 6}) = \pm \frac{π}{2}

$II :$ $Re (\frac{z - 8 i}{z + 6}) = 0$

Show that locus of $z$ in $I$ or $II$ lies on $x^{2} + y^{2} + 6 x - 8 y = 0$ . Hence, show that locus to $z$ can also be represented by $\frac{z - 8 i}{z + 6} + \frac{\bar{z} + 8 i}{\bar{z} + 6} = 0 .$ Further, if the locus of $z$ is expressed as $| z + 3 - 4 i | = R$ , then find $R$

MEDIUM

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 6 - Complex Numbers>Exercise-1>Q 33

Show that

z \bar{z} + (4 - 3 i) z + (4 + 3 i) \bar{z} + 5 = 0

represents circle. Hence find centre and radius.

HARD

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 6 - Complex Numbers>Exercise-1>Q 34

If

z_{1} & z_{2}

are two complex numbers & if

\arg \frac{z_{1} + z_{2}}{z_{1} - z_{2}} = \frac{π}{2}

but

|z_{1} + z_{2}| \neq |z_{1} - z_{2}|

then identify the figure formed by the points represented by

0, z_{1}, z_{2} & z_{1} + z_{2}

.

EASY

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 6 - Complex Numbers>Exercise-1>Q 36

When the polynomial

5 x^{3} + M x + N

is divided by

x^{2} + x + 1

, the remainder is

0

. Then find

M + N

.

EASY

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 6 - Complex Numbers>Exercise-1>Q 37

Show that

(1 - ω + ω^{2}) (1 - ω^{2} + ω^{4}) (1 - ω^{4} + ω^{8}) \dots \dots \dots .

to

2 n

factors

= 2^{2 n}

, where

ω

is the cube root of unity.

HARD

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 6 - Complex Numbers>Exercise-1>Q 38

Let

ω

is non-real root of

x^{3} = 1

.

(i) If $P = ω^{n}, (n \in N)$ and

$Q = ({}^{2 n}C_{0} + {}^{2 n}C_{3} + \dots \dots . .) + (^{2 n} C_{1} + {}^{2 n}C_{4} + \dots \dots \dots ω + ({}^{2 n}C_{2} + {}^{2 n}C_{5} + . . . .) ω^{2}$ , then find $\frac{P}{Q}$ .

(ii) If $P = 1 - \frac{ω}{2} + \frac{ω^{2}}{4} - \frac{ω^{3}}{8} \dots \dots$ upto $\infty$ terms and $Q = \frac{1 - ω^{2}}{2}$ , then find value of $P Q$ .