JEE Advanced Level Test: Complex Number- 2 - JEE MCQ

Let   

Using Componendo and Dividenode, we get

an imaginary number


= an imaginary number

= an imaginary number

 we get 

We have, zk = ωk where ω  



(by De Movire's theorem)
= cos π + i sin π = -1

Since, it given that
(a₁ +ib₁)(a₂ +ib₂) ....(a_n + ib_n) = A +iB

Taking conjugate throughout, we have

Multiplying (i) and (ii), we have

Since, ω³ = 1, hence

Option (a) :- for x = nπ
let n=1
so, x= 1 x π =π
therefore , sinx + icos2x
sinπ + icos 2π
0 + i= i
Now,
cosx - isin2x
cosπ - isin2π
or, -1 -0 = -1
hence no conjugate
Option (b) :- x = 0
sin0 + icos0
0+i= i
now ,
cos0 - isin0
= 1
hence no conjugate.
option (c) :- x = (n+1/2)π
if n= 1
x=3π/2
sin3π/2 + icos3π
-1 - i
Now,
cos3π/2 - isin3π
0-0= 0
hence no conjugate
Thus option D is correct

Since, α lies i III Quadrant and only tan is positive in this quadrant
⇒ |z| = -sec α

z₁ = (1, α) = cosα + isinα = e^iα  ......(i)
⇒ z₂ = (1, β) cosβ+isinβ = e^i/β  ......(ii)
⇒ z₃ =(1, γ). cosγ +isinγ =e^iγ   ......(iii)
Also, Z₁ +Z₂ +Z₃ = 0 

⇒ (cosα+cosβ+cosγ)+i(sinα+sinβ+sinγ) = 0
⇒ ccosα+cosβ+cosγ = 0
and sinα+sinβ+sinγ = 0 ... (iv) 

[(Using (i), (ii), (iii)] 
⇒ E = (cosα+cosβ+cosγ - i (sinα+sinβ+sinγ)
   [(Using (i)]

    (Triangle Inequality)
 ....(i)

       ....(ii)

Adding (i), (ii),(iii), we have 

⇒ αS = α + 2α² + 3α³ + ...+(n -1)α^n-1 +nαⁿ

On subtracting, we get

Since, α is nth root of unity, therfore
αⁿ = 1 

Since, 1, ω, ω² ..... ω^n-1 are the nth roots of unity, therefore
xⁿ - 1 = (x -1)(x - ω)(x - ω2⁾...(x - ω^n-1)

Putting x = 1 in above equation, we have

n,n + 1, n +2  are consecutive integers their sum is muttiple of 3, so 

According to Triangle Inequality, we have

Now,  i is an increasing function of |z| and |z| > 5
 will take its minimum value at z = 5
⇒ minimum value of 

Let z = 0 + ib, where b < 0. Then z is represented by a point on the negative direction of y-axis. 

According to Trianglr Inequality
 (given)

We have, α + β = -p,αβ = q
Now (ωα + ω²β)(ω²α + ωβ) = ω³α² +ω⁴αβ+ω²αβ + ω³β²
= α² +β² +(ω+ω²)αβ = α²+β² -αβ = (α + β)² -3αβ = p² -3q

z =| z | -3+ 2i ⇒ | z |²= (| z | -3)² + 4 = | z |² -6 | z | +9 + 4Þ | z | 13/6
=
Thus,  z = - 5/6 + 2i

⇒ z is purely real.
∴ Im (z) = 0

⇒ |z| = 1

As β ≠ 1 is afourth root of unity,
β4 = 1 ⇒ (1 - β) (1 + β + β² + β³) = 0


∴ z = 0

so that the equation becomes 1/z = 1z²

where ω (≠1) is a cube root of unity.