Complex numbers and quadratic equations is a segment of maths that deals with crucial theorems and concepts along with various formulae. It comprises of linear and quadratic equations along with roots related to the complex number's set (known as complex roots)..

**Q1.**If Î±,Î² be the roots of the equation u^2-2u+2=0 and if cotÎ¸= x + 1, then [( x + Î± )^n - (x + Î²)^n ]/[Î± - Î²] is equal to

Solution

(a)

u^2 - 2u + 2 = 0 ⇒ u = 1 ± i

⇒( (x + Î±)^n - ( x + Î²)^n ) /( Î± - Î²)

=( [ (cotÎ¸ - 1) + (1 + i) ]^n - [ ( cotÎ¸ - 1 ) + (1 - i) ]^n )/2i (∵cotÎ¸ - 1 = x)

=( (cosÎ¸ + isinÎ¸ )^n - (cosÎ¸ - isinÎ¸ )^n )/( sin^nÎ¸ 2i)

=( 2i sinnÎ¸ )/( sin^nÎ¸ 2i )

=sinnÎ¸ / sin^nÎ¸

(a)

u^2 - 2u + 2 = 0 ⇒ u = 1 ± i

⇒( (x + Î±)^n - ( x + Î²)^n ) /( Î± - Î²)

=( [ (cotÎ¸ - 1) + (1 + i) ]^n - [ ( cotÎ¸ - 1 ) + (1 - i) ]^n )/2i (∵cotÎ¸ - 1 = x)

=( (cosÎ¸ + isinÎ¸ )^n - (cosÎ¸ - isinÎ¸ )^n )/( sin^nÎ¸ 2i)

=( 2i sinnÎ¸ )/( sin^nÎ¸ 2i )

=sinnÎ¸ / sin^nÎ¸

**Q2.**If the cube roots of unity are 1 , Ï‰ , Ï‰^2, then the roots of the equation (x- 1)^3 + 8 = 0 are

Solution

(b)

( (x - 1)/( -2 ) )^3 = 1

⇒(x - 1)/ (-2) = 1 , Ï‰ , Ï‰^2

⇒x = -1 , 1 - 2Ï‰ , 1 , -2Ï‰^2

(b)

( (x - 1)/( -2 ) )^3 = 1

⇒(x - 1)/ (-2) = 1 , Ï‰ , Ï‰^2

⇒x = -1 , 1 - 2Ï‰ , 1 , -2Ï‰^2

**Q3.**Suppose A is a complex number and n∈N, such that A^n = [(A + 1)]^n = 1, then the least value of n is

Solution

(b)

Let A = x + iy . Given,

| A | = 1 ⇒ x^2 + y^2 = 1 and

| A + 1 | = 1

⇒( x + 1 )^2 + y^2 = 1

⇒x = -1/2 and y = ±√3/2

⇒A = Ï‰ or Ï‰^2

⇒(Ï‰)^n = (1 + Ï‰)^n = (-Ï‰^2 )^n

Therefore, n must be even and divisible by 3

(b)

Let A = x + iy . Given,

| A | = 1 ⇒ x^2 + y^2 = 1 and

| A + 1 | = 1

⇒( x + 1 )^2 + y^2 = 1

⇒x = -1/2 and y = ±√3/2

⇒A = Ï‰ or Ï‰^2

⇒(Ï‰)^n = (1 + Ï‰)^n = (-Ï‰^2 )^n

Therefore, n must be even and divisible by 3

**Q4.**If a,b,c,d are four consecutive terms of an increasing A.P. then the roots of the equation (x - a) (x - c) + 2(x - b) (x - d) = 0 are

Solution

(d)

Given that a < b < c < d. let

f(x) = (x - a)( x - c) + 2(x - b)(x - d)

⇒f(b) = (b - a)(b - c) < 0

and f(d) = (d - a)(d - c) > 0

Hence, f(x) = 0 has one root in (b , d). Also, f(a)f(c) < 0. So the other root lies in (a , c). Hence, roots of the equation are real and distinct

(d)

Given that a < b < c < d. let

f(x) = (x - a)( x - c) + 2(x - b)(x - d)

⇒f(b) = (b - a)(b - c) < 0

and f(d) = (d - a)(d - c) > 0

Hence, f(x) = 0 has one root in (b , d). Also, f(a)f(c) < 0. So the other root lies in (a , c). Hence, roots of the equation are real and distinct

**Q5.**If the equations ax^2 + bx + c = 0 and x^3 + 3x^2 + 3x + 2 = 0 have two common roots, then

Solution

(a) By observation x=-2 satisfies equation x^3 + 3x^2 + 3x + 2 = 0 then we have

(x + 2)( x^2 + x + 1) = 0

x^2 + x + 1 = 0 has non-real roots

Since non-real roots occur in conjugate pair,

x^2 + x + 1 = 0 and ax^2 + bx + c = 0 are identical

⇒a=b=c

(a) By observation x=-2 satisfies equation x^3 + 3x^2 + 3x + 2 = 0 then we have

(x + 2)( x^2 + x + 1) = 0

x^2 + x + 1 = 0 has non-real roots

Since non-real roots occur in conjugate pair,

x^2 + x + 1 = 0 and ax^2 + bx + c = 0 are identical

⇒a=b=c

**Q6.**If Î± and Î² , Î± and Î³ , Î± and Î´ are the roots of the equations ax^2 + 2bx + c = 0, 2bx^2 + cx + a = 0 and cx^2 + ax + 2b = 0, respectively, where a,b and c are positive real numbers, then Î± + Î±^2 =

Solution

(c)

Since Î± is root of all equations

aÎ±^2 + 2bÎ± + c = 0

2bÎ±^2 + cÎ± + Î± = 0

cÎ±^2 + aÎ± + 2b = 0

Adding we get ( a + 2b + c )( Î±^2 + Î± + 1 ) = 0

a + 2b + c ≠ 0 as a , b , c > 0

⇒ Î±^2 + Î± + 1 = 0 or Î±^2 + Î± = -1

(c)

Since Î± is root of all equations

aÎ±^2 + 2bÎ± + c = 0

2bÎ±^2 + cÎ± + Î± = 0

cÎ±^2 + aÎ± + 2b = 0

Adding we get ( a + 2b + c )( Î±^2 + Î± + 1 ) = 0

a + 2b + c ≠ 0 as a , b , c > 0

⇒ Î±^2 + Î± + 1 = 0 or Î±^2 + Î± = -1

**Q7.**If Î±,Î² are the roots of x^2 + px + q = 0 and x^2n + p^n x^n + q^n = 0 and if ( Î±/Î² ) , ( Î²/Î± ) are the roots of x^n + 1 + (x + 1)^n = 0, then n(∈N)

Solution

(c)

We have,

Î± + Î² =-p and Î±Î² = q (1)

Also, since Î± , Î² are the root of x^2n + p^n x^n + q^n = 0,

we have ,

Î±^2n + p^n Î±^n + q^n = 0 and Î²^2n + p^n Î²^n + q^n = 0

Subtracting the above relations, we get

(Î±^2n - Î²^2n ) + p^n (Î±^n - Î²^n ) = 0

∴ Î±^n + Î²^n = -p^n (2)

Given, Î±/Î² or Î²/Î± is a root of x^n + 1 + (x + 1)^n = 0. So,

( Î±/Î² )^n + 1 + [ (Î±/Î²) + 1 ]^n = 0

⇒( Î±^n + Î²^n ) + ( Î± + Î² )^n = 0

⇒ -p^n + (-p)^n = 0 [Using (1) and (2)]

It is possible only when n is even

(c)

We have,

Î± + Î² =-p and Î±Î² = q (1)

Also, since Î± , Î² are the root of x^2n + p^n x^n + q^n = 0,

we have ,

Î±^2n + p^n Î±^n + q^n = 0 and Î²^2n + p^n Î²^n + q^n = 0

Subtracting the above relations, we get

(Î±^2n - Î²^2n ) + p^n (Î±^n - Î²^n ) = 0

∴ Î±^n + Î²^n = -p^n (2)

Given, Î±/Î² or Î²/Î± is a root of x^n + 1 + (x + 1)^n = 0. So,

( Î±/Î² )^n + 1 + [ (Î±/Î²) + 1 ]^n = 0

⇒( Î±^n + Î²^n ) + ( Î± + Î² )^n = 0

⇒ -p^n + (-p)^n = 0 [Using (1) and (2)]

It is possible only when n is even

**Q8.**If z^2 + z |z| + |z^2 | = 0, then the locus of z is

Solution

(c)

z^2 + z|z| + |z|^2 = 0

⇒( z/|z| )^2 + z/|z| + 1 = 0

⇒z/|z| = Ï‰ , Ï‰^2 ⇒ z = Ï‰ |z| or z = Ï‰^2 |z|

⇒x + iy = |z| (( -1 )/2 + ( i√3 ) / 2) or x + iy =| z |((-1)/2 - (i√3)/2)

⇒x = -1/2 |z|, y = |z|√3/2 or x = -|z|/2, y =- (|z| √3)/2

⇒y + √3 x = 0 or y - √3 x = 0 ⇒ y^2 - 3x^2 = 0

(c)

z^2 + z|z| + |z|^2 = 0

⇒( z/|z| )^2 + z/|z| + 1 = 0

⇒z/|z| = Ï‰ , Ï‰^2 ⇒ z = Ï‰ |z| or z = Ï‰^2 |z|

⇒x + iy = |z| (( -1 )/2 + ( i√3 ) / 2) or x + iy =| z |((-1)/2 - (i√3)/2)

⇒x = -1/2 |z|, y = |z|√3/2 or x = -|z|/2, y =- (|z| √3)/2

⇒y + √3 x = 0 or y - √3 x = 0 ⇒ y^2 - 3x^2 = 0

**Q9.**If |z|=1 and w=(z-1)/(z+1) (where z≠-1), then Re(w) is

Solution

(a)

Since, | z | = 1 and w = (z - 1)/(z + 1)

⇒ z = (1 + w)/(1 - w)

⇒ | z |=( | 1 + w |)/(| 1 - w |)

⇒ | 1 - w | = | 1 + w | [ ∵ | z | = 1 ]

⇒ 1 + | w |^2 - 2 Re (w) = 1 + | w |^2 + 2 Re (w)

⇒ Re (w) = 0

(a)

Since, | z | = 1 and w = (z - 1)/(z + 1)

⇒ z = (1 + w)/(1 - w)

⇒ | z |=( | 1 + w |)/(| 1 - w |)

⇒ | 1 - w | = | 1 + w | [ ∵ | z | = 1 ]

⇒ 1 + | w |^2 - 2 Re (w) = 1 + | w |^2 + 2 Re (w)

⇒ Re (w) = 0

**Q10.**If the equation x^2 + ax + b = 0 has distinct real roots and x^2 + a| x | + b = 0 has only one real root, then which of the following is true