# Complex Number

Algebra

## 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| .

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

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

Sol. 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

Sol: AB/BC = √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.