JEE  >  Subjective Type Questions: Complex Numbers | JEE Advanced

# Subjective Type Questions: Complex Numbers | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE

## Document Description: Subjective Type Questions: Complex Numbers | JEE Advanced for JEE 2022 is part of Maths 35 Years JEE Main & Advanced Past year Papers preparation. The notes and questions for Subjective Type Questions: Complex Numbers | JEE Advanced have been prepared according to the JEE exam syllabus. Information about Subjective Type Questions: Complex Numbers | JEE Advanced covers topics like and Subjective Type Questions: Complex Numbers | JEE Advanced Example, for JEE 2022 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for Subjective Type Questions: Complex Numbers | JEE Advanced.

Introduction of Subjective Type Questions: Complex Numbers | JEE Advanced in English is available as part of our Maths 35 Years JEE Main & Advanced Past year Papers for JEE & Subjective Type Questions: Complex Numbers | JEE Advanced in Hindi for Maths 35 Years JEE Main & Advanced Past year Papers course. Download more important topics related with notes, lectures and mock test series for JEE Exam by signing up for free. JEE: Subjective Type Questions: Complex Numbers | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE
 1 Crore+ students have signed up on EduRev. Have you?

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 = ω
so that ω + ω2 + 1= 0 and ω3 =1.
Then xyz = (a + b) (aω2 + bω) (aω + bω2)
= (a + b) (a2ω3 + abω4 + abω+ 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 + b  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 z1+ z2+ z3= 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 ......, an – 1  are the n roots of unity, then show that (1 – a1)(1 – a2) (1 – a3) ....(1 – an – 1) = n (1984 - 2 Marks)

Ans.

Sol.  1, a1, a2, .... an – 1 are the n roots of unity. Clearly above n values are roots of eq. xn – 1 = 0

Therefore we must have (by factor theorem)

xn – 1 = (x – 1) (x – a1) (x – a2) ....(x – an – 1) ....(1)

= (x – a1) (x – a2) ....(x – an – 1) ....(2)

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

nxn – 1 = (x – a1) (x – a2) ... (x – an – 1) + (x – 1) (x – a2) ....(x – an – 1) +....+ (x – 1) (x – a1) ... (x – an – 2)

For x = 1, we get n = (1– a1) (1– a2) .... (1 – an – 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 – R, 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  z= 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)
⇒ 6y – 36 = x2 + y2 –14x – 12y + 40 + 36
⇒ x2 + y2 – 14x – 18y + 112 = 0
⇒ (x– 14x + 49) + (y– 18y + 81) =18
⇒ ( x - 7)+ (y - 9)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 =-
z– iz+ iz + 1= 0

or z2(z – i) + i (z – i) = 0

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

or     z2 = – i

∴ | z | = | i | = 1 or | z2 | = | z | = | – 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 r≤ and r2≤ 1

We have Z - W = (rcos θ1 -r2 cosθ2)

+i(rsin θ1 -r2 sinθ2)

|Z - W|= (r1 cos θ1 -rcosθ2) + (r1 sin θ1 -rsinθ2)2

= r12 cos2 θ1 + r2cos2 θ2 - 2 r1r2 cos θ1 cos θ2 +r1sin2θ1

+ r2 sin θ- 2 r1r2 sin θ1 sinθ2

= r12 (cos2 θ1 + sinθ1) +r22 (cos2 θ2 + sin2θ2)

- 2 r1r(cos θ1 cos θ2 + sin θ1 sinθ2)

= = r12 + r22 - 2 r1r2 cos ( θ12)

= (r- r2)+ 2r1r2[1 - cos ( θ12)]

= (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 zand 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 zin 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,  z1z= q

By rotation through a in anticlockwise direction

z= z1 e ....(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|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 zp+q - zp - zq + 1 = 0, where p, q are distinct primes. Show that either 1 + α + α2 + .... + αp - 1 = 0 or 1 + α + α2 + ... + αq - 1 = 0, but not both together. (2002 - 5 Marks)

Ans.

Sol. The given equation can be written as

(z–1) (z– 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 αp = 1 and if αq = 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|) (1–|z2|) < 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|< 2

⇒ a1 z + a2z2 + a3z3 + ... + anzn = 1

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

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

∴ Using its generalised form, we get

| a1 z + a2 z2  + az3 + ... + an zn |

≤ | a1 z | + | a2 z2 | + ... + | azn |

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

But given that | a| < 2 ∀ r= 1(1)

∴ 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 = α+ iα2,  β = β+ 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'

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 zby an ∠90° in anticlockwise sense, about centre z0 Thus, z2 –  z0 = (z1 – z0) eiπ/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

The document Subjective Type Questions: Complex Numbers | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE is a part of the JEE Course Maths 35 Years JEE Main & Advanced Past year Papers.
All you need of JEE at this link: JEE

## Maths 35 Years JEE Main & Advanced Past year Papers

132 docs|70 tests
 Use Code STAYHOME200 and get INR 200 additional OFF

## Maths 35 Years JEE Main & Advanced Past year Papers

132 docs|70 tests

### Top Courses for JEE

Track your progress, build streaks, highlight & save important lessons and more!

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

;