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 x^2 + x + 1 = 0, then the value of [( x + 1/x )]^2 + [(x^2 + 1/x^2 )]^2+⋯+[( x^27 + 1/x^27 )]^2 is

Solution

(d)

x^2 + x + 1 = 0 ⇒ x = Ï‰ or Ï‰^2

Let x = Ï‰. Then,

x + 1/x = Ï‰ + 1/Ï‰ = Ï‰ + Ï‰^2 = -1

x^2 + 1/x^2 = Ï‰^2 + 1/Ï‰^2 = Ï‰^2 + Ï‰ = -1

x^3 + 1/x^3 = Ï‰^3 + 1/Ï‰^3 = 2

x^4 + 1/x^4 = Ï‰^4 + 1/Ï‰^4 = Ï‰ + Ï‰^2 = -1,etc

∴ (x + 1/x)^2 + (x^2 + 1/x^2 )^2 + (x^3 + 1/x^3 )^2 +⋯+ (x^27 + 1/x^27 )^2

= [(x + 1/x)^2 + (x^2 + 1/x^2 )^2 + (x^4 + 1/x^4 )^2 +⋯+ (x^26 + 1/x^26 )^2 ] + [(x^3 + 1/x^3 )^2 + (x^6 + 1/x^6 )^2 + (x^9 + 1/x^9 ) +⋯+ (x^27 + 1/x^27 )^2 ]

= 18 + 9(2^2) = 54

(d)

x^2 + x + 1 = 0 ⇒ x = Ï‰ or Ï‰^2

Let x = Ï‰. Then,

x + 1/x = Ï‰ + 1/Ï‰ = Ï‰ + Ï‰^2 = -1

x^2 + 1/x^2 = Ï‰^2 + 1/Ï‰^2 = Ï‰^2 + Ï‰ = -1

x^3 + 1/x^3 = Ï‰^3 + 1/Ï‰^3 = 2

x^4 + 1/x^4 = Ï‰^4 + 1/Ï‰^4 = Ï‰ + Ï‰^2 = -1,etc

∴ (x + 1/x)^2 + (x^2 + 1/x^2 )^2 + (x^3 + 1/x^3 )^2 +⋯+ (x^27 + 1/x^27 )^2

= [(x + 1/x)^2 + (x^2 + 1/x^2 )^2 + (x^4 + 1/x^4 )^2 +⋯+ (x^26 + 1/x^26 )^2 ] + [(x^3 + 1/x^3 )^2 + (x^6 + 1/x^6 )^2 + (x^9 + 1/x^9 ) +⋯+ (x^27 + 1/x^27 )^2 ]

= 18 + 9(2^2) = 54

Solution

(c)

¯z + i¯w =0

⇒ z - iw = 0 (1)

⇒ z = iw

argzw = Ï€

⇒ argz + argw = Ï€

⇒ argz + arg[z/i] = Ï€ [Using (1)]

⇒ argz + argz - argi = Ï€

⇒ 2argz - Ï€/2 = Ï€

⇒ 2argz = 3Ï€/2

⇒ argz = 3Ï€/4

(c)

¯z + i¯w =0

⇒ z - iw = 0 (1)

⇒ z = iw

argzw = Ï€

⇒ argz + argw = Ï€

⇒ argz + arg[z/i] = Ï€ [Using (1)]

⇒ argz + argz - argi = Ï€

⇒ 2argz - Ï€/2 = Ï€

⇒ 2argz = 3Ï€/2

⇒ argz = 3Ï€/4

**Q4.**If a > 0 , b > 0 and c > 0 then the roots of the equation ax^2 + bx + c = 0

Solution

(c)

As a , b , c > 0 , so a , b , c should be real (note that other relation is not defined in the set of complex numbers). Therefore, the roots of equation are either real or complex conjugate Let Î±,Î² be the roots of ax^2 + bx + c = 0. Then,

Î± + Î² = -b/a = -ve and Î±Î² = c/a = +ve

Hence, either both Î±,Î² are –ve (if roots are real) or both Î±,Î² have –ve real part (if roots are complex conjugate)

(c)

As a , b , c > 0 , so a , b , c should be real (note that other relation is not defined in the set of complex numbers). Therefore, the roots of equation are either real or complex conjugate Let Î±,Î² be the roots of ax^2 + bx + c = 0. Then,

Î± + Î² = -b/a = -ve and Î±Î² = c/a = +ve

Hence, either both Î±,Î² are –ve (if roots are real) or both Î±,Î² have –ve real part (if roots are complex conjugate)

**Q5.**Which of the following is equal to ∛(-1)?

Solution

(b)

x = ∛(-1)

⇒x^3 = -1

⇒(-x)^3 = 1

⇒ -x = 1 , Ï‰ , Ï‰^2

⇒ x = -1 , -Ï‰ , -Ï‰^2

= -1 ,( 1 + √3 i )/2 , ( 1 - √3i )/2

= -1 ,( -√3 + i )/2i ,( √3 + i )/2i

=-1,( -√3 + √(-1) )/√(-4) ,( √3 + √(-1) )/√(-4)

(b)

x = ∛(-1)

⇒x^3 = -1

⇒(-x)^3 = 1

⇒ -x = 1 , Ï‰ , Ï‰^2

⇒ x = -1 , -Ï‰ , -Ï‰^2

= -1 ,( 1 + √3 i )/2 , ( 1 - √3i )/2

= -1 ,( -√3 + i )/2i ,( √3 + i )/2i

=-1,( -√3 + √(-1) )/√(-4) ,( √3 + √(-1) )/√(-4)

**Q6.**The interval of a for which the equation tan^2[x - (a - 4 )] tanx + 4 - 2a = 0 has at least one solution ∀ x ∈ [0 , Ï€/4]

Solution

(b)

tan(x) = ( a - 4 - √( ( a - 4 )^2 - 4(4 - 2a ) ) )/2

= ( a - 4 - a )/2 = a - 2 ,-2

∴ tanx = a - 2 (∵ tanx ≠ -2)

∵ x ∈ [0 , Ï€/4]

∴ 0 ≤ a -2 ≤ 1

⇒ 2 ≤ a ≤3

(b)

tan(x) = ( a - 4 - √( ( a - 4 )^2 - 4(4 - 2a ) ) )/2

= ( a - 4 - a )/2 = a - 2 ,-2

∴ tanx = a - 2 (∵ tanx ≠ -2)

∵ x ∈ [0 , Ï€/4]

∴ 0 ≤ a -2 ≤ 1

⇒ 2 ≤ a ≤3

**Q7.**Which of the following represents a point in an Argand plane, equidistant from the roots of the equation (z + 1)^4 = 16z^4 ?

Solution

(c)

( (z + 1 )/z )^4 = 16

⇒ ( z + 1 )/z = ±2 , ±2i

The roots are 1 , -1/3 ,( -1/5 - (2/5)i) and ( -1/5 + (2/5)i)

Note that (-1/3,0) and (1, 0) are equidistant from (1/3, 0) and since it lies on the perpendicular bisector of AB, it will be equidistant from A and B also

(c)

( (z + 1 )/z )^4 = 16

⇒ ( z + 1 )/z = ±2 , ±2i

The roots are 1 , -1/3 ,( -1/5 - (2/5)i) and ( -1/5 + (2/5)i)

Note that (-1/3,0) and (1, 0) are equidistant from (1/3, 0) and since it lies on the perpendicular bisector of AB, it will be equidistant from A and B also

**Q8.**If Î± , Î² , Î³ are such that Î± + Î² + Î³ = 2 ,Î±^2 + Î²^2 + Î³^2 = 6, Î±^3 + Î²^3 + Î³^3 = 8, then Î±^4 + Î²^4 + Î³^4 is

Solution

(a)

We have,

(Î± + Î² + Î³)^2 = Î±^2 + Î²^2 + Î³^2 + 2(Î²Î³ + Î³Î± + Î±Î²)

⇒ 4 = 6 + 2 (Î²Î³ + Î³Î± + Î±Î²)

⇒ Î²Î³ + Î³Î± + Î±Î² = -1

Also, Î±^3 + Î²^3 + Î³^3 - 3Î±Î²Î³ =(Î± + Î² + Î³)(Î±^2 + Î²^2 + Î³^2 - Î²Î³ - Î³Î± - Î±Î²)

⇒ 8 - 3Î±Î²Î³ = 2(6 + 1)

⇒ 3Î±Î²Î³ = 8 - 14 = -6 or Î±Î²Î³ = -2

Now,

(Î±^2 + Î²^2 + Î³^2 )^2 = ∑(Î±^4) + 2∑(Î±^2 Î²^2 )

= ∑(Î±^4 + 2)[(∑(Î±Î²)^2 - 2Î±Î²Î³(∑(Î±) ) )]

⇒ ∑(Î±^4) = 36 - 2[ (-1)^2 -2(-2)(2) ] = 18

(a)

We have,

(Î± + Î² + Î³)^2 = Î±^2 + Î²^2 + Î³^2 + 2(Î²Î³ + Î³Î± + Î±Î²)

⇒ 4 = 6 + 2 (Î²Î³ + Î³Î± + Î±Î²)

⇒ Î²Î³ + Î³Î± + Î±Î² = -1

Also, Î±^3 + Î²^3 + Î³^3 - 3Î±Î²Î³ =(Î± + Î² + Î³)(Î±^2 + Î²^2 + Î³^2 - Î²Î³ - Î³Î± - Î±Î²)

⇒ 8 - 3Î±Î²Î³ = 2(6 + 1)

⇒ 3Î±Î²Î³ = 8 - 14 = -6 or Î±Î²Î³ = -2

Now,

(Î±^2 + Î²^2 + Î³^2 )^2 = ∑(Î±^4) + 2∑(Î±^2 Î²^2 )

= ∑(Î±^4 + 2)[(∑(Î±Î²)^2 - 2Î±Î²Î³(∑(Î±) ) )]

⇒ ∑(Î±^4) = 36 - 2[ (-1)^2 -2(-2)(2) ] = 18

**Q9.**The minimum value of |a + bÏ‰ + cÏ‰^2 |, where a , b and c are all not equal integers and Ï‰( ≠ 1) is a cube root of unity, is

Solution

(c)

Let x=|a + bÏ‰ + cÏ‰^2 |

⇒ x^2 = (a^2 + b^2 + c^2 - ab - bc - ca)

⇒ x^2 = 1/2{(a - b)^2 + (b - c)^2 + (c - a)^2} …(i)

Since a , b , c are all integers but not all simultaneously equal

⇒ If a = b, then a ≠ c and b ≠ c

As, difference of integers = integer

⇒ (b - c)^2 ≥ 1

[as minimum difference of two consecutive integers is ( ±1 )]

Also, (c - a)^2 ≥ 1

∴ From Eq. (i),

x^2 = 1/2[(a - b)^2 + (b - c)^2 + (c - a)^2 ] ≥ 1/2[0 + 1 + 1]

⇒ x^2 ≥ 1

Hence, minimum value of x is 1

(c)

Let x=|a + bÏ‰ + cÏ‰^2 |

⇒ x^2 = (a^2 + b^2 + c^2 - ab - bc - ca)

⇒ x^2 = 1/2{(a - b)^2 + (b - c)^2 + (c - a)^2} …(i)

Since a , b , c are all integers but not all simultaneously equal

⇒ If a = b, then a ≠ c and b ≠ c

As, difference of integers = integer

⇒ (b - c)^2 ≥ 1

[as minimum difference of two consecutive integers is ( ±1 )]

Also, (c - a)^2 ≥ 1

∴ From Eq. (i),

x^2 = 1/2[(a - b)^2 + (b - c)^2 + (c - a)^2 ] ≥ 1/2[0 + 1 + 1]

⇒ x^2 ≥ 1

Hence, minimum value of x is 1

Solution

(a)

|z_1 | = |z_2 | = |z_3 | = 1

Hence, the circumcentre of triangle is origin. Also, centroid ( z_1 + z_2 + z_3 )/3 = 0, which coincides with the circumcentre. So the triangle is equilateral. Since radius is 1, length of side is a=√3. Therefore, the area of the triangle is (√3/4) a^2 = (3√3/4)

(a)

|z_1 | = |z_2 | = |z_3 | = 1

Hence, the circumcentre of triangle is origin. Also, centroid ( z_1 + z_2 + z_3 )/3 = 0, which coincides with the circumcentre. So the triangle is equilateral. Since radius is 1, length of side is a=√3. Therefore, the area of the triangle is (√3/4) a^2 = (3√3/4)