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

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

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

(C) form a regular polygon of unit circum-radius .

(D) are non-collinear.

Sol. Clearly, roots are 1, α, α2 , . . . αn-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

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

(C) arg(z) = ± 2π/3

(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…

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)

(C) f(1) = g(1) ≠ 0

(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

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

(C) $\alpha -\bar{a}$ 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

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)

(C) 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

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

(A) |z + z| = 1/2

(B) z + z = 1

(C) |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

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

(C) 2

(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

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

(A) Re (z) > 0

(B) Re (z) < 0

(C) Re (z) > 2

(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

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

(C) a circle passing through the origin

(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

Q: 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

(A) greater than 2/3

(B) less than 2/3

(C) greater than |cos θ1| + |cos θ2| + … + |cos θn|

(D) less than |cos θ1| + |cos θ2| + … + |cos θn|

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

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

3 ≤ |z|n + |z|n-1 + … + 1

Since |tan θi| ≤ 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.