Complex Number – QuantumStudy

Roots of the equation x^n –1 = 0, n ∈I,

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Q: Roots of the equation xⁿ –1 = 0, n ∈ I,

(A) are collinear (B) lie on a circle.

(D) are non-collinear.

Sol. Clearly, roots are 1, α, α² , . . . α^n-1 , where $\large \alpha = cos\frac{2\pi}{n} + i sin\frac{2\pi}{n}$ .

The distance of the complex numbers represented by these roots from origin is 1 i.e. all these points lie on a circle.

⇒ They are non-collinear.

⇒ They form a regular polygon of unit circum-radius.

Hence (B), (C) and (D) are the correct answers.

Let ‘z’ be a complex number and ‘a’ be a real parameter such that z^2 + az + a^2 = 0, then

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Q: Let ‘z’ be a complex number and ‘a’ be a real parameter such that z² + az + a² = 0, then

(A) locus of z is a pair of straight lines

(B) locus of z is a circle

(D) |z| =|a|

Click to See Answer :

Ans: (A),(C) ,(D)

Sol: z² + az + a² = 0

⇒ z = aw, aw² ( where ‘w’ is non real root of cube unity )

⇒ locus of z is a pair of straight lines

and arg (z) =arg(a) + arg(w) or arg(a) + arg(w²)

⇒ arg(z) = ± 2π/3

also, |z| = |a||w| or |a| |w²| ⇒ |z| = |a|

Hence (A), (C) and (D) are the correct answers.

If f(x) and g(x) are two polynomials such that the polynomial $h(x) = x f(x^3) + x^2 g(x^6)$ is divisible…

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Q: If f(x) and g(x) are two polynomials such that the polynomial $h(x) = x f(x^3) + x^2 g(x^6)$ is divisible by x² +x +1 , then

(A) f(1) = g(1)

(B) f(1) = – g( 1)

(D) f(1) = -g(1) ≠ 0

Click to See Answer :

Ans: (A) & (B)

Sol: Roots of x² + x +1 = 0 are complex cube roots of unity,

so h(w) = h(w²) =0

⇒ w f(1) + w² g(1) = 0 and w² f(1) + wg(1) = 0

⇒ f(1) = g(1) = 0.

Hence (A) and (B) are the correct answers.

If the reflection of the line $\bar{a}z + a \bar{z}= 0 $ in the real axis is $\bar{\alpha}z + \alpha \bar{z}= 0 $ in the simplest form

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Q: If the reflection of the line $\bar{a}z + a \bar{z}= 0 $ in the real axis is $\bar{\alpha}z + \alpha \bar{z}= 0 $ in the simplest form, then

(A) α + a is purely real

(B) $\bar{\alpha} – a $ is purely real

(D) $ \bar{\alpha} -\bar{a}$ is purely imaginary

Ans: (B) & (C)

Solution: Let a = a₁ + ia₂ and z = x + iy, then

$\bar{a}z + a \bar{z}= 0 $

⇒ a₁ x + a₂ y = 0

or , $\large y = (-\frac{a_1}{a_2})x $

Its reflection in the real axis (x-axis) is $\large y = (\frac{a_1}{a_2})x $

or, a₁ x – a₂ y = 0

i.e. $\large (\frac{a + \bar{a}}{2}) (\frac{z + \bar{z}}{2}) – (\frac{a – \bar{a}}{2i}) (\frac{z – \bar{z}}{2i})= 0 $

So, $\large \bar{\alpha} = a $ and $\large \alpha = \bar{a}$

Hence (B) and (C) are the correct answers.

If A(z1), B(z2) and C(z3) be the vertices of a triangle ABC in which ∠ABC =π/4 and AB/BC =√2 , then the value of z2 is equal to

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Q: If A(z₁), B(z₂) and C(z₃) be the vertices of a triangle ABC in which ∠ABC = π/4 and AB/BC =√2 , then the value of z₂ is equal to

(A) z₃ + i(z₁ + z₃)

(B) z₃ – i(z₁ – z₃)

(D) None of these

Click to See Answer :

Ans: (C)

Sol: $\displaystyle \frac{AB}{BC} = \sqrt{2}$

Considering the rotation about ‘B’ ,we get,

$\large \frac{z_1 -z_2}{z_3 -z_2} = \frac{|z_1-z_2|}{|z_3 -z_2|} e^{i\pi/4}$

$\large = \frac{AB}{BC} e^{i\pi/4} = \sqrt{2} (\frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}})$

= 1 + i

⇒ z₁ – z₂ = ( 1 + i) (z₃ – z₂)

⇒ z₁ – (1 + i) z₃ = z₂( 1 –1 – i) = – iz₃

⇒ z₂ = iz₁ – i( 1+i) z₃ = z₃ + i(z₁ – z₃)

Hence (C) is the correct answer.

If |z| = min {|z – 1| , |z + 1|} , then

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Q: If |z| = min {|z – 1|, |z + 1|}, then

(A) |z + z^–| = 1/2

(B) z + z^– = 1

(D) none of these

Click to See Answer :

Ans: (C)

Sol: Sol. |z| = min {|z – 1|, |z + 1|} = |z – 1| for Re (z) > 0

But |z| = |z – 1| ⇒ Re (z) = 1/2 ,

Similarly, |z| = min {|z – 1|, |z + 1|} = |z + 1|, for Re (z) < 0

⇒ |z| = |z + 1| ⇒ Re (z) = -1/2

where Re(z) =1/2 , z + z^– = 1 , |z + z^–| = 1

when Re (z) = 0 ⇒ z + z^– = 0 and z + z^– = 0

Hence the argument is |z + z^– | = 1

Hence (C) is the correct answer.

If the equation |z-z1|^2 +|z-z2|^2 = k, represents the equation of a circle, where z1= 2+3i, z2=4+3i are

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Q: If the equation |z-z₁|² +|z-z₂|² = k , represents the equation of a circle, where z₁ = 2 + 3i, z₂ = 4 + 3i are the extremities of a diameter, then the value of k is

(A) 1/ 4

(B) 4

(D) None of these

Click to See Answer :

Ans: (B)

Sol: As z₁ and z₂ are the extremities of the diameter,

⇒ |z – z₁|² +|z – z₂|² = |z₁ – z₂|²

⇒ k = |z₁ – z₂|² = | 2 + 3i – 4 – 3i|² = |-2|² = 4

Hence (B) is the correct answer.

The inequality |z-4| < |z-2|represents the region given by

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Q: The inequality |z-4| < |z-2|represents the region given by

(A) Re (z) > 0

(B) Re (z) < 0

(D) None of these

Click to See Answer :

Ans: (D)

Sol: |z – 4| < | z-2|

⇒ | z- 4|² < | z – 2|²

⇒ ( z-4) (z^– -4) < ( z-2) (z^– -2)

⇒ z z^–– 4(z + z^–) +16 < z – 2( z + z^–) + 4

⇒ 2( z + z^– ) > 12

⇒ 4Re (z) > 12

⇒ Re (z) > 3

Hence (D) is the correct answer.

The complex numbers z = x + iy which satisfy the equation

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Q: The complex numbers z = x + iy which satisfy the equation $\large |\frac{z-5i}{z+5i}|$ lie on

(A) the x-axis

(B) the straight line y = 5

(D) None of these

Sol. $\large |\frac{z-5i}{z+5i}| = 1 $

⇒ |z – 5i| = |z + 5i|

⇒ z would lie on the right bisector of the line segment connecting the points 5i and – 5i . Thus z would lie on the x-axis.

Hence (A) is the correct answer.

The distances of the roots of the equation tan θ0 zn + tan θ1 zn-1 + … + tan θn = 3 from z = 0, where θ0, θ1, θ2, …, θn ∈ [0 , π/4] satisfy

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Q: The distances of the roots of the equation tan θ0 zⁿ + tan θ1 z^n-1 + … + tan θn = 3 from z = 0, where θ0, θ1, θ2, …, θn ∈ [0 , π/4] satisfy

(A) greater than 2/3

(B) less than 2/3

Sol: 3 = |tan θ0 zⁿ + tan θ1 z^n-1 + .. + tan θn|

3 ≤ |tan θ0| |z|ⁿ + |tan θ1| |z|^n-1 + … + |tan θn|

3 ≤ |z|ⁿ + |z|^n-1 + … + 1

Since |tan θ_i| ≤ 1

3 < 1 + |z| + |z|² + … ∞

$\large 3 < \frac{1}{1-|z|}$

3 – 3 |z| < 1

– 3 |z| < 1 – 3 = – 2

⇒ 3 |z| > 2

⇒ |z| > 2/3.

Hence (A) is the correct answer.