JEE Advanced (Subjective Type Questions): Complex Numbers | Chapter-wise Tests for JEE Main & Advanced PDF Download

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Q. 1. Express   in the form x + iy. (1978)

Ans. 

Sol. 

= 

which is of the form X + iY.

Q. 2. If x = a + b, y  = aγ + bβ and z = aβ + bγ where γ and b are the complex cube roots of unity, show that xyz = a³ + b³. (1978)

Ans. Sol. As b and γ are the complex cube roots of unity therefore,
let β = ω and γ = ω² 
so that ω + ω² + 1= 0 and ω³ =1.
Then xyz = (a + b) (aω² + bω) (aω + bω²)
= (a + b) (a²ω³ + abω⁴ + abω² + b²ω³)
= (a + b) (a² + abω + abω² + b²) (using ω³ = 1)
= (a + b) (a² + ab(ω + ω² ) + b²)
= (a + b) (a² – ab + b² ) (using ω+ ω² = –1)
= a³ + b³   Hence proved.

Q. 3. If x + iy =  prove that (x² + y²)² =             (1979)

Ans.

 Sol. Given x + iy =

⇒ (x + iy)² ....(1)

Taking conjugate on both sides, we get

(x - iy)^{2 =}     ....(2)

Multiply (1)  and (2),  we get

(x² + y²)^{2 =} 

Q. 4. Find the real values of x and y for which the following equation is satisfied    (1980)

Ans. 

Sol. 

⇒ (4 + 2i) x – 6i – 2 + (9 – 7i) y + 3i – 1= 10i
⇒ (4x + 9y – 3) + (2x – 7y – 3) i = 10i
⇒ 4x + 9y – 3 = 0 and 2x – 7y – 3 = 10
On solving these two, we get x = 3, y = – 1
 

Q. 5. Let the complex number z₁, z₂ and z₃ be the vertices of an equilateral triangle. Let z₀ be the  circumcentre of the triangle.
 Then prove that z₁² + z₂² + z₃² = 3z₀². (1981 - 4 Marks)

Ans. 

Sol. 

Let  us consider the equilateral Δ with each side of length 2a and having two of its vertices on x-axis namely A (–a,0) and B (a, 0), then third vertex C will clearly lie on y-axis s.t.
OC = 2a sin 60°= ∴ C has the co-ordinates (0,).
Now in the form of complex numbers if A, B and C are represented by z₁, z₂, z₃ then z₁ = - a ; z₂ = a ; z₃= a₃i As in an equilateral Δ, centriod and circumcentre coincide, we get

Circumcentre, 

⇒ 

Now,   

and  

∴  Clearly  

Q. 6. Prove that the complex numbers z₁, z₂ and the origin form an equilateral triangle only if z₁² + z₂² –z₁z₂ = 0. (1983 - 3 Marks)

 Ans. Sol. We know that if z₁, z₂, z₃   are vertices of an equilateral Δ then

Here 

We get  

⇒ - ( z₁ - z₂)= z₁z₂

⇒ - z₁² - z₂² + 2 z₁ z₂= z₁z₂ ⇒ z₁² + z₂² - z₁z₂= 0.

Q. 7. If 1, a₁, a₂ ......, a_{n – 1}  are the n roots of unity, then show that (1 – a₁)(1 – a₂) (1 – a₃) ....(1 – a_{n – 1}) = n (1984 - 2 Marks)

Ans. 

Sol.  1, a₁, a₂, .... a_{n – 1} are the n roots of unity. Clearly above n values are roots of eq. x_{n – 1} = 0

Therefore we must have (by factor theorem)

x_{n – 1} = (x – 1) (x – a₁) (x – a₂) ....(x – a_{n – 1}) ....(1)

⇒  = (x – a₁) (x – a₂) ....(x – a_{n – 1}) ....(2)

Differentiating both sides of eq. (1), we get

nxn – 1 = (x – a₁) (x – a₂) ... (x – a_{n – 1}) + (x – 1) (x – a₂) ....(x – a_{n – 1}) +....+ (x – 1) (x – a₁) ... (x – a_{n – 2})

For x = 1, we get n = (1– a₁) (1– a₂) .... (1 – a_{n – 1})

[All the terms except first contain (x – 1) and hence become zero for x = 1] Proved.

Q. 8. Show that the area of the triangle on the Argand diagram formed by the complex numbers z, iz and z + iz is  (1986 - 2½ Marks)

Ans.

 Sol. Let A = z = x + iy, B = iz = –  y + ix,

C = z + iz = (x – y) + i (x + y)

Now, area of ΔABC =

Operating   R₂ – R₁ , R₃ – R₁,   we  get

 Δ = 

 |x(-y - x) +y (x - y)|

=  |- xy -x² + xy-y²| = |-x²-y²|

= |x²+y²|= |z²|           Hence Proved.

Q. 9. Let Z₁ = 10 + 6i and Z₂ = 4 + 6i. If Z is any complex number such that the argument of  , then prove that |Z – 7 – 9i| =  . (1990 -  4 Marks)

Ans. Sol. We are given that z₁ = 10 + 6i  and  z₂ = 4 + 6i

Also arg

⇒ arg  (z – z₁) – arg (z – z₂) = =  NOTE THIS STEP

⇒ arg ((x+iy) - (10 + 6i)) -arg((x+iy) - (4 + 6i)) =

⇒ arg [(x - 10) +i (y - 6)] -arg [( x - 4) + i(y - 6)] =

⇒ tan^-1 - tan^-1 =

⇒ tan^-1 =

⇒ = tan 

⇒ (x – 4 – x + 10) (y – 6) = (x – 4) (x – 10) + (y – 6)² 
⇒ 6y – 36 = x² + y² –14x – 12y + 40 + 36
⇒ x² + y² – 14x – 18y + 112 = 0
⇒ (x² – 14x + 49) + (y² – 18y + 81) =18
⇒ ( x - 7)² + (y - 9)²= ()² 
⇒ ( x + iy) - (7 + 9i)=
⇒ z - (7 + 9i)= .              Hence Proved.

Q. 10. If iz³ + z² – z + i = 0,  then show that | z | = 1. (1995 -  5 Marks)

Ans.

 Sol. Dividing through out by i and knowing that 1/i = -i we get i =-
z³ – iz² + iz + 1= 0

or z²(z – i) + i (z – i) = 0    

as   1= – i² or (z – i) (z² + i) = 0 ∴z = i    

or     z² = – i

∴ | z | = | i | = 1 or | z² | = | z |²  = | – i | = 1

⇒ | z | =1

Hence in either case | z | = 1

Q. 11. If |Z|≤ 1, |W |≤ 1, show that
 |Z -W|² ≤ (|Z|- |W|)² +( Arg Z- ArgW)² (1995 -  5 Marks)

Ans. Sol. Let Z = r₁ (cos θ₁ +i sinθ₁)

and W = r₂ (cos θ₂ +i sinθ₂)

We have | Z | = r₁, | W | = r₂, Arg  Z= θ₁  and

Arg W = θ₂

Since | Z | ≤ 1, |W |≤ 1, it follows that r₁ ≤ and r₂≤ 1

We have Z - W = (r_l cos θ₁ -r₂ cosθ₂)

+i(r_l sin θ₁ -r₂ sinθ²)

|Z - W|² = (r₁ cos θ₁ -r₂ cosθ₂) + (r₁ sin θ₁ -r₂ sinθ₂)²

= r₁² cos² θ₁ + r₂² cos² θ₂ - 2 r₁r₂ cos θ₁ cos θ₂ +r₁² sin²θ₁

+ r₂ sin θ₂ - 2 r₁r₂ sin θ₁ sinθ₂

= r₁² (cos² θ₁ + sin² θ₁) +r₂² (cos² θ₂ + sin²θ₂)

- 2 r₁r₂ (cos θ₁ cos θ₂ + sin θ₁ sinθ₂)

= = r₁² + r₂² - 2 r₁r² cos ( θ₁ -θ₂)

= (r₁ - r₂)² + 2r₁r₂[1 - cos ( θ₁ -θ₂)]

= (r1 - r2) ² + 4 r₁r₂ sin² 

= |r₁ - r₂|² + 4 r₁r₂

≤ |r₁ - r₂|² + 4 [ ∵ r₁, r₂ ≤ 1]

But | sin θ| ≤| θ| ∀ θ∈R               

                                                  NOTE THIS STEP

Therefore,

Thus   + (Arg Z- ArgW)²

12. Find all non-zero complex numbers Z satisfying  (1996 - 2 Marks)

Ans. 

Sol.  Let z = x + iy then
⇒ x – iy = i(x² – y² + 2ixy)

⇒ x – iy = i(x² – y²) – 2xy

⇒ x (1 + 2y) = 0 ; x² – y² + y = 0

⇒ x = 0 or y = -⇒ x = 0,  y = 0,  1

or 

For non zero complex number  z

 x = 0, y = 1; 

∴   

Q. 13. Let z₁ and z₂ be roots of the equation z²+pz+q = 0, where the coefficients p and q may be complex numbers. Let A and  B represent z₁ and z₂ in the complex plane. If ∠AOB = a ≠ 0 and OA = OB, where O is the origin, prove that

p² = 4q cos2            (1997 - 5 Marks)

Ans. Sol. z² + pz + q = 0

z₁+ z₂= –  p,  z₁z₂ = q

By rotation through a in anticlockwise direction

z₂ = z₁ e^iα ....(1)

Add 1 in both sides to get z₁ + z₂ = – p

∴ 

or 

On squaring (z₁ + z₂)² 

 =

or p² = 4q  cos²

Q. 14. For complex numbers z and w, prove that |z|² w-|w|² z = z –w if and only if z = w or z . (1999 - 10 Marks) 

 Ans. 

Sol. Given that z and w are two complex numbers.
To prove    |z|² w –|w|² z = z – w ⇔ z = w or z

First let us consider

|z|² w –|w|² z = z – w ....(1)

⇒ z (1 +|w|² = w ( 1 +|z|²)

⇒  =  a real number

⇒  

⇒                 ....(2)

Again from equation (1),

       (Using equation (2))

⇒  ⇒    or z = w 

Conversely if  z = w then

L.H.S.  of  (1) =|w|²  w –|w|²  w = 0.

R.H.S.  of  (1) =  w – w = 0

∴ (1) holds

Also if  z  then

L.H.S. of  (1) 

= z – w = R .H.S. Hence proved.

Q. 15. Let a complex number α, α ≠ 1 , be a root of the equation z^p+q - z^p - z^q + 1 = 0, where p, q are distinct primes. Show that either 1 + α + α² + .... + α^{p - 1} = 0 or 1 + α + α² + ... + α^{q - 1} = 0, but not both together. (2002 - 5 Marks)

Ans. 

Sol. The given equation can be written as

(z^p –1) (z^q – 1) = 0

∴ z = (1)^1/p     or         (1)^1/q ....(1)

where p and q are distinct prime numbers.
Hence both the equations will have distinct roots and as

z ≠ 1, both will not be simultaneously zero for any value of z given by equations in (1)

NOTE THIS STEP

Also  
or 

Because of (1) either α^p = 1 and if α^q = 1 but not both simultaneously as p and q are distinct primes.

Q. 16. If z₁ and z₂ are two complex numbers such taht |z₁| < 1< |z₂| then prove that   (2003 - 2 Marks)

Ans. 

Sol. Given that|z₁|<1<|z₂|

Then  < 1 is true

if <|z₁– z₂|is true

if  <  |z₁– z₂|² is true

if  <  (z₁ - z₂)  is true

if  <  (z₁ - z₂)

if 

- is true

if  1+|z₁|²|z₂|²<|z₁|²+|z₂|²      is true

if  (1 –|z₁|² ) (1–|z₂|² ) < 0 is true.

which is obviously true as|z₁|< 1<|z₂|

⇒|z₁|² < 1<|z₂|²

⇒|1–|z₁|² > 0 and  (1–|z₂|²) < 0              Hence proved.

Q. 17. Prove that there exists no complex number z such that and  = 1    where |ar| < 2. (2003 - 2 Marks)

Ans. 

Sol. Let us consider,  where | a_r |< 2

⇒ a₁ z + a₂z² + a₃z³ + ... + a_nzⁿ = 1

⇒ |a₁z + a₂z² + a₃z³ + ...+ a_nzⁿ| = 1                   ....(1)

But we know that | z₁ + z₂ | ≤ | z₁ |+ |z₂|

∴ Using its generalised form, we get

| a₁ z + a₂ z²  + a₃ z³ + ... + a_n zⁿ |

≤ | a₁ z | + | a₂ z² | + ... + | a_n zⁿ |

⇒ 1 ≤ | a₁| | z | + | a₂ |  |z² | + | a₃ | | z³ | + ... + | a_n | |zⁿ | (Using eqn (1))

But given that | a_r | < 2 ∀ r= 1(1)ⁿ 

∴ 1< 2 [ | z | + | z |² + | z |³ + ... + | z |ⁿ ] [Using | zⁿ | = | z |ⁿ ]

⇒ 1< 2  ⇒  

⇒ 2 [ | z | – | z |n+1]  > 1– | z |      (∵ 1– | z | > 0 as | z | < 1/3)

⇒  [|z| -⇒ 

⇒  [|z|  ⇒  [|z| >

which is a contradiction as given that   [|z| <

∴ There exist no such complex number.

Q. 18. Fin d the centre an d radius of circle given by = k, k ≠ 1

where, z = x + iy, α = α₁ + iα₂,  β = β₁ + iβ₂ (2004 - 2 Marks) 

Ans. Sol. We are given that

 = k ⇒ | z -a|= k |z -b|

Let pt. A represents complex number α and B that of  β, and P represents z. then | z – α | = k | z – β |

⇒ z  is the complex number whose distance from A is k times its distance from B. i.e. PA = k PB

⇒ P divides AB in the ratio k : 1 internally or externally (at P').

Then  and 

Now through PP' there can pass a number of circles, but with given data we can find radius and centre of that circle for which PP' is diameter.
And hence then centre = mid. point of PP'

Also radius

Q. 19. If one the vertices of the square circumscribing the circle  |z – 1|  = . Find the other vertices of thesquare. (2005 - 4 Marks)

Ans. Sol.  The given circle is |z - 1|= where z₀=1 is the centre and  is radius of circle. z₁ is one of the vertex of square inscribed in the given circle.

Clearly z₂ can be obtained by rotating z₁ by an ∠90° in anticlockwise sense, about centre z₀ Thus, z₂ –  z₀ = (z₁ – z₀) e^iπ/2

or z₂ - 1 = (2 + i- 1)i ⇒ z₂ =i -+1

z₂ = (1 - )+i

Again rotating z₂ by 90° about z₀ we get

z₃ – z₀ = (z₂ – z₀) i

⇒ z₃ -1= [(1- ) + i -1] i = - i-1 ⇒ z₃ =-i

and similarly 1 = (- i - 1) i =-i

⇒ z₄ = (  + 1)-i

Thus the remaining vertices are

(1 - ) + i, - i , ( + 1)-i