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.**The complex numbers z = x + iy which satisfy the equation |( z - 5i )/(z + 5i )| = 1 lie on

Solution

(a)

We know that |z - z_1 | = |z - z_2 |. Then locus of z is the line, which is a perpendicular bisector of line segment joining z_1 and z_2

Hence,

z = x + iy

⇒ |z - 5i| = |z + 5i|

Therefore, z remains equidistant from z_1 = 5i and z_2 = 5i. Hence, z lies on perpendicular bisector of line segment joining z_1 and z_2, which is clearly the real axis or y = 0

Alternative solution:

|( z - 5i )/( z + 5i )| = 1

⇒ | x + iy - 5i| = |x + iy + 5i|

⇒ | x + (y - 5)i | = |x + (y + 5) i|

⇒ x^2 + (y - 5)^2 = x^2 + (y + 5)^2

⇒ x^2 + y^2 - 10y + 25 = x^2 + y^2 + 10y + 25

⇒ 20y = 0

⇒ y = 0

(a)

We know that |z - z_1 | = |z - z_2 |. Then locus of z is the line, which is a perpendicular bisector of line segment joining z_1 and z_2

Hence,

z = x + iy

⇒ |z - 5i| = |z + 5i|

Therefore, z remains equidistant from z_1 = 5i and z_2 = 5i. Hence, z lies on perpendicular bisector of line segment joining z_1 and z_2, which is clearly the real axis or y = 0

Alternative solution:

|( z - 5i )/( z + 5i )| = 1

⇒ | x + iy - 5i| = |x + iy + 5i|

⇒ | x + (y - 5)i | = |x + (y + 5) i|

⇒ x^2 + (y - 5)^2 = x^2 + (y + 5)^2

⇒ x^2 + y^2 - 10y + 25 = x^2 + y^2 + 10y + 25

⇒ 20y = 0

⇒ y = 0

**Q2.**If Î± and Î² (Î± < Î²) are the roots of the equation x^2 + bx + c = 0, where c<0

Solution

(b)

Here D = b^2 - 4c > 0 because c < 0 < b. So, roots are real and unequal. Now,

Î± + Î² = -b < 0 and Î±Î² = c < 0

Therefore, one root is positive and the other root is negative, the negative root being numerically bigger. As Î±<Î², so Î± is the negative root while Î² is the positive root. So, |Î±| > Î² and Î± < 0 < Î² < |Î±|

(b)

Here D = b^2 - 4c > 0 because c < 0 < b. So, roots are real and unequal. Now,

Î± + Î² = -b < 0 and Î±Î² = c < 0

Therefore, one root is positive and the other root is negative, the negative root being numerically bigger. As Î±<Î², so Î± is the negative root while Î² is the positive root. So, |Î±| > Î² and Î± < 0 < Î² < |Î±|

**Q3.**All the values of m for which both the roots of the equation x^2 - 2mx + m^2 - 1 = 0 are greater than -2 but less than 4, lie in the interval

Solution

(c)

The given equation is

x^2 - 2mx + m^2 - 1 = 0

⇒ (x - m)^2 - 1 = 0

⇒ (x - m + 1)(x - m - 1) = 0

⇒ x = m - 1 , m + 1

From given condition,

m - 1 > -2 and m + 1 < 4

⇒ m > -1 and m < 3

Hence, -1 < m < 3

(c)

The given equation is

x^2 - 2mx + m^2 - 1 = 0

⇒ (x - m)^2 - 1 = 0

⇒ (x - m + 1)(x - m - 1) = 0

⇒ x = m - 1 , m + 1

From given condition,

m - 1 > -2 and m + 1 < 4

⇒ m > -1 and m < 3

Hence, -1 < m < 3

**Q4.**Two towns A and B are 60 km apart. A school is to be built to serve 150 students in town A and 50 students in town B. If the total distance to be travelled by all 200 students is to be as small as possible, then the school be built at

Solution

(c)

Let the distance of the school from A be x. Therefore, the distance of the school from B is 60 - x. The total distance covered by 200 students is

[150x + 50( 60 - x )] = [ 100x + 3000 ]

This is minimum when x = 0. Hence, the school should be at town A

(c)

Let the distance of the school from A be x. Therefore, the distance of the school from B is 60 - x. The total distance covered by 200 students is

[150x + 50( 60 - x )] = [ 100x + 3000 ]

This is minimum when x = 0. Hence, the school should be at town A

**Q5.**If z = [ (√3/2) + i/2 ]^5 + [ (√3/2) - i/2 ]^5, then

Solution

(b)

z = (√3/2 + i/2)^5 + ( √3/2 - i/2)^5

= ( cos[Ï€/6] + isin[Ï€/6] )^5 + ( cos[Ï€/6]- isin[Ï€/6] )^5

= (cos[5Ï€/6] + isin[5Ï€/6]) + (cos[5Ï€/6] - isin[5Ï€/6] )

= 2cos[5Ï€/6]

= -√3

⇒ Re(z) < 0 and Im(z) = 0

Alternative solution:

z = ¯z_1 + ¯z_2

Where

( √3/2 + i/2 )^5

⇒ z is real

⇒ Im(z) = 0

(b)

z = (√3/2 + i/2)^5 + ( √3/2 - i/2)^5

= ( cos[Ï€/6] + isin[Ï€/6] )^5 + ( cos[Ï€/6]- isin[Ï€/6] )^5

= (cos[5Ï€/6] + isin[5Ï€/6]) + (cos[5Ï€/6] - isin[5Ï€/6] )

= 2cos[5Ï€/6]

= -√3

⇒ Re(z) < 0 and Im(z) = 0

Alternative solution:

z = ¯z_1 + ¯z_2

Where

( √3/2 + i/2 )^5

⇒ z is real

⇒ Im(z) = 0

**Q6.**Let p and q be real numbers such that p ≠ 0 , p^3 ≠ q and p^3 ≠ -q. If Î± and Î² are non-zero complex numbers satisfying Î± + Î² = -p and Î±^3 + Î²^3 = q, then a quadratic equation having Î±/Î² and Î²/Î± as its roots is

Solution

(b)

Sum of roots = Î±/Î² + Î²/Î± = ( Î±^2 + Î²^2 )/Î±Î² and product = 1

Given, Î± + Î² = -p and Î±^3 + Î²^3 = q

⇒ (Î± + Î² )( Î±^2 - Î±Î² + Î²^2 ) = q

∴ Î±^2 + Î²^2 - Î±Î² = (-q)/p ...(i)

And (Î± + Î²)^2 = p^2

⇒ Î±^2 + Î²^2 + 2Î±Î² = p^2

From Eqs. (i) and (ii), we get

Î±^2 + Î²^2 = ( p^3 - 2q )/3p

And Î±Î² = ( p^3 + q )/3p

∴ Required equation is

x^2 - ( ( p^3 - 2q ) x)/( ( p^3 + q ) ) + 1 = 0

⇒ ( p^3 + q )x^2-( p^3 - 2q )x + ( p^3 + q ) = 0

(b)

Sum of roots = Î±/Î² + Î²/Î± = ( Î±^2 + Î²^2 )/Î±Î² and product = 1

Given, Î± + Î² = -p and Î±^3 + Î²^3 = q

⇒ (Î± + Î² )( Î±^2 - Î±Î² + Î²^2 ) = q

∴ Î±^2 + Î²^2 - Î±Î² = (-q)/p ...(i)

And (Î± + Î²)^2 = p^2

⇒ Î±^2 + Î²^2 + 2Î±Î² = p^2

From Eqs. (i) and (ii), we get

Î±^2 + Î²^2 = ( p^3 - 2q )/3p

And Î±Î² = ( p^3 + q )/3p

∴ Required equation is

x^2 - ( ( p^3 - 2q ) x)/( ( p^3 + q ) ) + 1 = 0

⇒ ( p^3 + q )x^2-( p^3 - 2q )x + ( p^3 + q ) = 0

**Q7.**Let p and q be roots of the equation x^2 - 2x + A = 0 and let r and s be the roots of the equation x^2 - 18x + B = 0. If p < q < r < s are in arithmetic progression, then the values of A and B are

Solution

(d)

Let the four numbers in A.P. be p = a - 3d , q = a - d , r = a + d , s = a + 3d. Therefore,

p + q = 2 , r + s = 18

Given that pq = A , rs = B

∴ p + q + r + s = 4a = 20

⇒ a = 5

Now, p + q = 2 ⇒ 10 - 4d = 2

r + s = 18 ⇒ 10 + 4d = 18

∴ d = 2

Hence, the numbers are -1 , 3 , 7 , 11

pq = A = -3 , rs = B = 77

(d)

Let the four numbers in A.P. be p = a - 3d , q = a - d , r = a + d , s = a + 3d. Therefore,

p + q = 2 , r + s = 18

Given that pq = A , rs = B

∴ p + q + r + s = 4a = 20

⇒ a = 5

Now, p + q = 2 ⇒ 10 - 4d = 2

r + s = 18 ⇒ 10 + 4d = 18

∴ d = 2

Hence, the numbers are -1 , 3 , 7 , 11

pq = A = -3 , rs = B = 77

**Q8.**If Î± and Î² are the roots of the equation x^2 + px + q = 0, and Î±^4 and Î²^4 are the roots of x^2 - rx + q = 0, then the roots of x^2 - 4qx + 2q^2 - r = 0 are always

Solution

(d)

Î±,Î² are roots of x^2 + px + q = 0. Hence,

Î± + Î² = -p br Î±Î² = q

Now,

Î±^4 , Î²^4 are roots x^2 - px + q = 0. Hence,

Î±^4 + Î²^4 = r , Î±^4 Î²^4 = q

Now, for equation x^2 - 4qx + 2q^2 - r = 0, product of roots is

2q^2 - r = 2(Î±Î²)^2 - (Î±^4 + Î²^4)

= - (Î±^2 - Î²^2 )^2

< 0

As product of roots is negative, so the roots must be real

(d)

Î±,Î² are roots of x^2 + px + q = 0. Hence,

Î± + Î² = -p br Î±Î² = q

Now,

Î±^4 , Î²^4 are roots x^2 - px + q = 0. Hence,

Î±^4 + Î²^4 = r , Î±^4 Î²^4 = q

Now, for equation x^2 - 4qx + 2q^2 - r = 0, product of roots is

2q^2 - r = 2(Î±Î²)^2 - (Î±^4 + Î²^4)

= - (Î±^2 - Î²^2 )^2

< 0

As product of roots is negative, so the roots must be real

**Q9.**The shaded region, where

P ≡ (-1 , 0) , Q ≡ (-1 + √2 , √2)

R≡(-1 + √2 , -√2) , S ≡ (1 , 0) is represented by

Solution

(a)

Here, |PQ| = |PS| = |PR| = 2

∴ Shaded part represents the external part of circle having centre ( -1 , 0 ) and radius 2 As we know equation of circle having centre z_0 and radius r, is

|z - z_0 | = r

∴ |z — ( -1 + 0i )| > 2

⇒ |z + 1| > 2 …(i)

Also, argument of z + 1 with represent to positive direction of x-axis is Ï€/4

∴ arg[(z + 1) ≤ Ï€/4 ]

And argument of z + 1 in a] = anti-clockwise direction is –Ï€/4.

∴ -Ï€/4 ≤ arg(z + 1)

⇒|arg(z + 1) | ≤ Ï€/4

(a)

Here, |PQ| = |PS| = |PR| = 2

∴ Shaded part represents the external part of circle having centre ( -1 , 0 ) and radius 2 As we know equation of circle having centre z_0 and radius r, is

|z - z_0 | = r

∴ |z — ( -1 + 0i )| > 2

⇒ |z + 1| > 2 …(i)

Also, argument of z + 1 with represent to positive direction of x-axis is Ï€/4

∴ arg[(z + 1) ≤ Ï€/4 ]

And argument of z + 1 in a] = anti-clockwise direction is –Ï€/4.

∴ -Ï€/4 ≤ arg(z + 1)

⇒|arg(z + 1) | ≤ Ï€/4

**Q10.**Number of values of a for which equations x^3 + ax + 1 = 0 and x^4 + ax^2 + 1 = 0 have a common root

Solution

(b)

Given equations are

x^3 + ax + 1 = 0

or x^4 + ax^2 + x = 0 (1)

and x^4 + ax^2 + 1 = 0 (2)

From (1) – (2), we get x = 1. Thus, x = 1 is the common roots.

Hence,

1 + a + 1 = 0

⇒ a = -2

(b)

Given equations are

x^3 + ax + 1 = 0

or x^4 + ax^2 + x = 0 (1)

and x^4 + ax^2 + 1 = 0 (2)

From (1) – (2), we get x = 1. Thus, x = 1 is the common roots.

Hence,

1 + a + 1 = 0

⇒ a = -2