# JEE Mathematics Handwritten Notes PDF for Revision 2023

## JEE Mathematics Handwritten Notes PDF for Revision 2023

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### JEE Main Maths Complex Numbers & Quadratic Equations Questions With Solutions

Question 1: If (1 + i) (1 + 2i) (1 + 3i) ….. (1 + ni) = a + ib, then what is 2 * 5 * 10….(1 + n2) is equal to?

Solution:

We have (1 + i) (1 + 2i) (1 + 3i) ….. (1 + ni) = a + ib …..(i)

(1 − i) (1 − 2i) (1 − 3i) ….. (1 − ni) = a − ib …..(ii)

Multiplying (i) and (ii),

we get 2 * 5 * 10 ….. (1 + n2) = a2 + b2

Question 2: If z is a complex number, then the minimum value of |z| + |z − 1| is ______.

Solution:

First, note that |−z|=|z| and |z1 + z2| ≤ |z1| + |z2|

Now |z| + |z − 1| = |z| + |1 − z| ≥ |z + (1 − z)|

= |1|

= 1

Hence, minimum value of |z| + |z − 1| is 1.

Question 3: For any two complex numbers z1 and z2 and any real numbers a and b; |(az1 − bz2)|2 + |(bz1 + az2)|2 = ___________.

Solution:

|(az1 − bz2)|2 + |(bz1 + az2)|2

= (a2 + b2) (|z1|2 + |z2|2)

Question 4: Find the complex number z satisfying the equations

Solution:

We have

,
Let z = x + iy, then

⇒ 3|z − 12| = 5 |z − 8i|

3 |(x − 12) + iy| = 5 |x + (y − 8) i|

9 (x − 12)2 + 9y2 = 25×2 + 25 (y − 8)2 ….(i) and

⇒ |z − 4| = |z − 8|

|x − 4 + iy| = |x − 8 + iy|

(x − 4)2 + y2 = (x − 8)2 + y2

⇒ x = 6

Putting x = 6 in (i), we get y2 − 25y + 136 = 0

y = 17, 8

Hence, z = 6 + 17i or z = 6 + 8i

Question 5: If z1 = 10 + 6i, z2 = 4 + 6i and z is a complex number such that

, then the value of |z − 7 − 9i| is equal to _________.
Solution:

Given numbers are z1 = 10 + 6i, z2 = 4 + 6i and z = x + iy

amp [(x − 10) + i (y − 6) (x − 4) + i (y − 6)] = π / 4

12y − y2 − 72 + 6y = x2 − 14x + 40 …..(i)

Now |z − 7 −9i| = |(x − 7) + i (y − 9)|

From (i), (x2 − 14x + 49) + (y2 − 18y + 81) = 18

(x − 7)2 + (y − 9)2 = 18 or

[(x − 7)2 + (y − 9)2]½ = [18]½ = 3√2
|(x − 7) + i (y − 9)| = 3√2 or

|z − 7 −9i| = 3√2.

Question 6: Suppose z1, z2, z3 are the vertices of an equilateral triangle inscribed in the circle |z| = 2. If z1 = 1 + i√3, then find the values of z3 and z2.

Solution:

One of the numbers must be a conjugate of z1 = 1 + i√3 i.e. z2 = 1 − i√3 or z3 = z1 ei2π/3 and

z2 = z1 e−i2π/3 , z3 = (1 + i√3) [cos (2π / 3) + i sin (2π / 3)] = −2

Question 7: If cosα + cos β + cos γ = sin α + sin β + sin γ = 0 then what is the value of cos 3α + cos 3β + cos 3γ?

Solution:

cos α + cos β + cos γ = 0 and sin α + sin β + sin γ = 0

Let a = cos α + i sin α; b = cos β + i sin β and c = cos γ + i sin γ.

Therefore, a + b + c = (cosα + cosβ + cosγ) + i (sinα + sinβ + sinγ) = 0 + i0 = 0

If a + b + c = 0, then a3 + b3 + c3 = 3abc or

(cosα + isina)3 + (cosβ + isinβ)3 + (cosγ + isinγ)3

= 3 (cosα + isinα) (cosβ + isinβ) (cosγ + isinγ)

⇒ (cos3α + isin3α) + (cos3β + isin3β) + (cos3γ + isin3γ)

= 3 [cos (α + β + γ) + i sin (α + β + γ)] or cos 3α + cos 3β + cos 3γ

= 3 cos (α + β + γ).

### JEE Main Maths , Relations And Functions Questions With Solutions

Question 8: If the cube roots of unity are 1, ω, ω2, then find the roots of the equation (x − 1)3 + 8 = 0.

Solution:

(x − 1)3 = −8 ⇒ x − 1 = (−8)1/3

x − 1 = −2, −2ω, −2ω2

x = −1, 1 − 2ω, 1 − 2ω2

Question 9: If 1, ω, ω2, ω3……., ωn−1 are the n, nth roots of unity, then (1 − ω) (1 − ω2) …..

(1 − ωn − 1) = ____________.

Solution:

Since 1, ω, ω2, ω3……., ωn−1 are the n, nth roots of unity, therefore, we have the identity

= (x − 1) (x − ω) (x − ω2) ….. (x − ωn−1) = xn − 1 or

(x − ω) (x − ω2)…..(x − ωn−1) = xn−1 / x−1

= xn−1 + xn−2 +….. + x + 1

Putting x = 1 on both sides, we get

(1 − ω) (1 − ω2)….. (1 − ωn−1) = n

Question 10: If a = cos (2π / 7) + i sin (2π / 7), then the quadratic equation whose roots are α = a + a2 + a4 and β = a3 + a5 + a6 is _____________.

Solution:

a = cos (2π / 7) + i sin (2π / 7)

a7 = [cos (2π / 7) + i sin (2π / 7)]7

= cos 2π + i sin 2π = 1 …..(i)

S = α + β = (a + a2 + a4) + (a3 + a5 + a6)

S = a + a2 + a3 + a4 + a5 + a6

or

P = α * β = (a + a2 + a4) (a3 + a5 + a6)

= a4 + a6 + a7 + a5 + a7 + a8 + a7 + a9 + a10

= a4 + a6 + 1 + a5 + 1 + a + 1 + a2 + a3 (From eqn (i)]

= 3+(a + a2 + a3 + a4 + a5 + a6)

= 3 + S = 3 − 1 = 2 [From (ii)]

Required equation is, x2 − Sx + P = 0

x2 + x + 2 = 0.

Question 11: Let z1 and z2 be nth roots of unity, which are ends of a line segment that subtend a right angle at the origin. Then n must be of the form ____________.

Solution:

11/n = cos [2rπ / n] + i sin [2r π / n]

Let z1 = [cos 2r1π / n] + i sin [2r1π / n] and z2 = [cos 2r2π / n] + i sin [2r2π / n].

Then ∠Z1 O Z2 = amp (z1 / z2) = amp (z1) − amp (z2)

= [2 (r1 − r2)π] / [n]

= π / 2

(Given) n = 4 (r1 − r2)

= 4 × integer, so n is of the form 4k.

Question 12: (cos θ + i sin θ)4 / (sin θ + i cos θ)5 is equal to ____________.

Solution:

(cos θ + i sin θ)4 / (sin θ + i cos θ)5

= (cos θ + i sin θ)4 / i5 ([1 / i] sin θ + cos θ)5

= (cosθ + i sin θ)4 / i (cos θ − i sin θ)5

= (cos θ + i sin θ)4 / i (cos θ + i sin θ)−5 (By property) = 1 / i (cos θ + i sin θ)9

= sin(9θ) − i cos (9θ).

Question 13: Given z = (1 + i√3)100, then find the value of Re (z) / Im (z).

Solution:

Let z = (1 + i√3)

r = √[3 + 1] = 2 and r cosθ = 1, r sinθ = √3, tanθ = √3 = tan π / 3 ⇒ θ = π / 3.

z = 2 (cos π / 3 + i sin π / 3)

z100 = [2 (cos π / 3 + i sin π / 3)]100

= 2100 (cos 100π / 3 + i sin 100π / 3)

= 2100 (−cos π / 3 − i sin π / 3)

= 2100(−1 / 2 −i √3 / 2)

Re(z) / Im(z) = [−1/2] / [−√3 / 2] = 1 / √3.

Question 14: If x = a + b, y = aα + bβ and z = aβ + bα, where α and β are complex cube roots of unity, then what is the value of xyz?

Solution:

If x = a + b, y = aα + bβ and z = aβ + bα, then xyz = (a + b) (aω + bω2) (aω2 + bω),where α = ω and β = ω2 = (a + b) (a2 + abω2 + abω + b2)

= (a + b) (a2− ab + b2)

= a3 + b3

Question 15: If ω is an imaginary cube root of unity, (1 + ω − ω2)7 equals to ___________.

Solution:

(1 + ω − ω2)7 = (1 + ω + ω2 − 2ω2)7

= (−2ω2)7

= −128ω14

= −128ω12ω2

= −128ω2