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 α,β be the roots of the equation u^2-2u+2=0 and if cotθ= x + 1, then [( x + α )^n - (x + β)^n ]/[α - β] is equal to
Solution
(a)
u^2 - 2u + 2 = 0 ⇒ u = 1 ± i
⇒( (x + α)^n - ( x + β)^n ) /( α - β)
=( [ (cotθ - 1) + (1 + i) ]^n - [ ( cotθ - 1 ) + (1 - i) ]^n )/2i (∵cotθ - 1 = x)
=( (cosθ + isinθ )^n - (cosθ - isinθ )^n )/( sin^nθ 2i)
=( 2i sinnθ )/( sin^nθ 2i )
=sinnθ / sin^nθ
(a)
u^2 - 2u + 2 = 0 ⇒ u = 1 ± i
⇒( (x + α)^n - ( x + β)^n ) /( α - β)
=( [ (cotθ - 1) + (1 + i) ]^n - [ ( cotθ - 1 ) + (1 - i) ]^n )/2i (∵cotθ - 1 = x)
=( (cosθ + isinθ )^n - (cosθ - isinθ )^n )/( sin^nθ 2i)
=( 2i sinnθ )/( sin^nθ 2i )
=sinnθ / sin^nθ
Q2.If the cube roots of unity are 1 , ω , ω^2, then the roots of the equation (x- 1)^3 + 8 = 0 are
Solution
(b)
( (x - 1)/( -2 ) )^3 = 1
⇒(x - 1)/ (-2) = 1 , ω , ω^2
⇒x = -1 , 1 - 2ω , 1 , -2ω^2
(b)
( (x - 1)/( -2 ) )^3 = 1
⇒(x - 1)/ (-2) = 1 , ω , ω^2
⇒x = -1 , 1 - 2ω , 1 , -2ω^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 = ω or ω^2
⇒(ω)^n = (1 + ω)^n = (-ω^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 = ω or ω^2
⇒(ω)^n = (1 + ω)^n = (-ω^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 α and β , α and γ , α and δ 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 α + α^2 =
Solution
(c)
Since α is root of all equations
aα^2 + 2bα + c = 0
2bα^2 + cα + α = 0
cα^2 + aα + 2b = 0
Adding we get ( a + 2b + c )( α^2 + α + 1 ) = 0
a + 2b + c ≠ 0 as a , b , c > 0
⇒ α^2 + α + 1 = 0 or α^2 + α = -1
(c)
Since α is root of all equations
aα^2 + 2bα + c = 0
2bα^2 + cα + α = 0
cα^2 + aα + 2b = 0
Adding we get ( a + 2b + c )( α^2 + α + 1 ) = 0
a + 2b + c ≠ 0 as a , b , c > 0
⇒ α^2 + α + 1 = 0 or α^2 + α = -1
Q7.
If α,β are the roots of x^2 + px + q = 0 and x^2n + p^n x^n + q^n = 0 and if ( α/β ) , ( β/α ) are the roots of x^n + 1 + (x + 1)^n = 0, then n(∈N)
Solution
(c)
We have,
α + β =-p and αβ = q (1)
Also, since α , β are the root of x^2n + p^n x^n + q^n = 0,
we have ,
α^2n + p^n α^n + q^n = 0 and β^2n + p^n β^n + q^n = 0
Subtracting the above relations, we get
(α^2n - β^2n ) + p^n (α^n - β^n ) = 0
∴ α^n + β^n = -p^n (2)
Given, α/β or β/α is a root of x^n + 1 + (x + 1)^n = 0. So,
( α/β )^n + 1 + [ (α/β) + 1 ]^n = 0
⇒( α^n + β^n ) + ( α + β )^n = 0
⇒ -p^n + (-p)^n = 0 [Using (1) and (2)]
It is possible only when n is even
(c)
We have,
α + β =-p and αβ = q (1)
Also, since α , β are the root of x^2n + p^n x^n + q^n = 0,
we have ,
α^2n + p^n α^n + q^n = 0 and β^2n + p^n β^n + q^n = 0
Subtracting the above relations, we get
(α^2n - β^2n ) + p^n (α^n - β^n ) = 0
∴ α^n + β^n = -p^n (2)
Given, α/β or β/α is a root of x^n + 1 + (x + 1)^n = 0. So,
( α/β )^n + 1 + [ (α/β) + 1 ]^n = 0
⇒( α^n + β^n ) + ( α + β )^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| = ω , ω^2 ⇒ z = ω |z| or z = ω^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| = ω , ω^2 ⇒ z = ω |z| or z = ω^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