## JEE Mathematics Handwritten Notes PDF for Revision 2023

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JEE Mathematics Handwritten Notes PDF for Revision 2023 | Free PDFs of Topper’s Notes . If you are a student of class 11 or 12th and want to crack JEE exam, then given JEE Mathematics Handwritten Notes PDF for Revision 2023 are the best resources to do this. As we know that MATHS is the basic subjects in JEE entrance exam which is held by NTA National test agency JEE (UG), or National Eligibility cum Entrance Test (Graduation). JEE exam is used for admission in IIT and NIT collages as well as alternative or traditional Engineering and Technical in India. Download JEE Mathematics Handwritten Notes PDF for Revision 2023 | Free PDFs of Topper’s Notes from the given links of Google drive.

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JEE Mathematics Handwritten Notes PDF for Revision 2023 he subject of MATHS will help you in preparing for your final exams. Your final exams in MATHS will be held in March or April. These MATHS pdf notes are short, neat and clean, colored, regularly updated, and according to the syllabus set by the CBSE board following NCERT guidelines. The notes are very useful for competitive exams like JEE, IIT, and ACT (American College Testing), etc…. Not only our Biology short notes are suitable for CBSE board students, but also for other state board students. If you want to get high marks in MATHS then you must study from these notes. To download JEE Mathematics Handwritten Notes PDF for Revision 2023 please read this article till the end.

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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).

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### 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

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