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Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 10 Ordinary Differential Equations Ex 10.2

Question 1.
Express each of the following physical statements in the form of differential equation.
(i) Radium decays at a rate proportional to the amount Q present.
Solution:
Radium decays at a rate proportional to the amount Q present.

(ii) The population P of a city increases at a rate proportional to the product of population and to the difference between 5,00,000 and the population.
Solution:
Rate of change of P with respect to ‘t’ is \(\frac{d \mathrm{P}}{d t}\)
Product of population and the difference between 50,000 and the population is P (50,000 – P)

(iii) For a certain substance, the rate of change of vapor pressure P with respect to temperature T is proportional to the vapor pressure and inversely proportional to the square of the temperature.
Solution:

(iv) A saving amount pays 8% interest per year, compounded continuously. In addition, the income from another investment is credited to the amount continuously at the rate of ₹ 400 per year.
Solution:
Let ‘x’ be the amount invested.

Question 2.
Assume that a spherical rain drop evaporates at a rate proportional to its surface area. Form a differential equation involving the rate of change of the radius of the rain drop.
Solution:

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Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 10 Ordinary Differential Equations Ex 10.1

Question 1.
For each of the following differential equations, determine its order, degree (if exists)

Solution:
Order = 1,
Degree = 1

(ii)
Solution:

(iii)
Solution:

(iv)
Solution:

(v)
Solution:

(vi)
Solution:

(vii)
Solution:

(viii)
Solution:
Order = 2

(ix)
Solution:

(x)
Solution:

Order = 1,
degree = Not exist

Samacheer Kalvi 12th Maths Solutions Chapter 10 Ordinary Differential Equations Ex 10.1 Additional Problems

Question 1.
Find the order and degree of the following differential equations:

Solution:
Order = 1,
Degree = 1

(ii) y’ + y² + y³ = 0
Solution:
Order = 1,
Degree = 1

(iii)
Solution:

(iv)
Solution:

∴ Order = 2
Degree = 2

(v)
Solution:

(vi)
Solution:

(vii)
Solution:

(viii)
Solution:

(ix)
Solution:

(x) sin x(dx + dy) = cos x(dx – dy)
Solution:
Order = 1;
Degree = 1

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Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 2 Complex Numbers Ex 2.8

Question 1.
If to ω ≠ 1 is a cube root of unity, then show that \(\frac{a+b \omega+c \omega^{2}}{b+c \omega+a \omega^{2}}+\frac{a+b \omega+c \omega^{2}}{c+a \omega+b \omega^{2}}=1\)
Solution:
Since ω is a cube root of unity, we have ω³ = 1 and 1 + ω + ω² = 0

Question 2.
Show that \(\left(\frac{\sqrt{3}}{2}+\frac{i}{2}\right)^{5}+\left(\frac{\sqrt{3}}{2}-\frac{i}{2}\right)^{5}=-\sqrt{3}\)
Solution:

Question 3.
Find the value of \(\left(\frac{1+\sin \frac{\pi}{10}+i \cos \frac{\pi}{10}}{1+\sin \frac{\pi}{10}-i \cos \frac{\pi}{10}}\right)^{10}\)
Solution:

Question 4.
If 2 cos α = x + \(\frac{1}{x}\) and 2 cos β = y + \(\frac{1}{y}\), show that

Solution:
(i) 2 cos α = x + \(\frac{1}{x}\)
⇒ 2 cos α = \(\frac{x^{2}+1}{x}\)
⇒ 2x cos α = x² + 1
⇒ x² – 2x cos α + 1 = 0

Question 5.
Solve the equation z³ + 27 = 0
Solution:
z³ + 27 = 0
⇒ z³ = -27
⇒ z³ = 3³(-1)
⇒ z = 3(-1)^1/3

Question 6.
If ω ≠ 1 is a cube root of unity, show that the roots of the equation (z – 1)³ + 8 = 0 are -1, 1 – 2ω, 1 – 2ω²
Solution:
(z – 1)³ + 8 = 0
⇒ (z – 1 )³ = -8

The roots are -1, 1 – 2ω, 1 – 2ω²

Question 7.
Find the value of \(\sum_{k=1}^{8}\left(\cos \frac{2 k \pi}{9}+i \sin \frac{2 k \pi}{9}\right)\)
Solution:
\(\sum_{k=1}^{8}\left(\cos \frac{2 k \pi}{9}+i \sin \frac{2 k \pi}{9}\right)\)
We know that 9th roots of unit are 1, ω, ω², ……., ω⁸
Sum of the roots:
1 + ω + ω² + …. + ω⁸ = 0 ⇒ ω + ω² + ω³ + …… + ω⁸ = -1

The sum of all the terms \(\sum_{k=1}^{8}\left(\cos \frac{2 k \pi}{9}+i \sin \frac{2 k \pi}{9}\right)\) = -1

Question 8.
If ω ≠ 1 is a cube root of unity, show that
(i) (1 – ω + ω²)⁶ + (1 + ω – ω²)⁶ = 128
(ii) (1 + ω)(1 + ω²)(1 + ω⁴)(1 + ω⁸)……(1 + ω²ⁿ) = 1
Solution:
(i) ω is a cube root of unity ω³ = 1; 1 + ω + ω² = 0
(1 – ω + ω²)⁶ + (1 + ω – ω²)⁶
= (-ω – ω)⁶ + (-ω² – ω²)⁶
= (-2ω)⁶ + (-2ω²)⁶
= (-2)⁶(ω⁶ + ω¹²)
= (64)(1 + 1)
= 128
(ii) (1 + ω)(1 + ω²)(1 + ω⁴)(1 + ω⁸) …… (1 + ω²ⁿ)
= (1 + ω)(1 + ω²)(1 + ω⁴)(1 + ω⁸) ……. 2n factors
= (-ω²)(-ω)(-ω²)(-ω) …… 2n factors
= ω³. ω³
= 1

Question 9.
If z = 2 – 2i, find the rotation of z by θ radians in the counter clockwise direction about the origin when
(i) θ = \(\frac{\pi}{3}\)
(ii) θ = \(\frac{2 \pi}{3}\)
(iii) θ = \(\frac{3 \pi}{2}\)
Solution:
(i) z = 2 – 2i = 2 (1 – i) = r(cos θ + i sin θ)
\(r=\sqrt{x^{2}+y^{2}}=2 \sqrt{1+1}=2 \sqrt{2}\)
\(\alpha=\tan ^{-1}=\left|\frac{y}{x}\right|=\tan ^{-1}|1|=\frac{\pi}{4}\)
(1 – i) lies in IV quadrant
θ = -α = \(-\frac{\pi}{4}\)
\(\Rightarrow z=2 \sqrt{2}\left[\cos \left(\frac{-\pi}{4}\right)+i \sin \left(\frac{-\pi}{4}\right)\right]\)
z is rotated by θ = \(\frac{\pi}{3}\) in the counter clock wise direction.

(ii) z is rotated by θ = \(\frac{2 \pi}{3}\) in the counter clockwise direction.

(iii) z is rotated by θ = \(\frac{3 \pi}{2}\) in the counter clockwise direction.

Question 10.
Prove that the values of \(\sqrt[4]{-1} \text { are } \pm \frac{1}{\sqrt{2}}(1 \pm i)\)
Solution:

The roots are \(\pm \frac{1}{\sqrt{2}}(1 \pm i)\)

Samacheer Kalvi 12th Maths Solutions Chapter 2 Complex Numbers Ex 2.8 Additional Problems

Question 1.

Solution:

Question 2.

Solution:

Question 3.
Prove that: (1 + i)⁴ⁿ and (1 + i)^{4n + 2} are real and purely imaginary respectively.
Solution:
Let z = 1 + i

= 2i (-1)ⁿ which is purely imaginary.

Question 4.

Solution:

Question 5.
If a = cos 2α + i sin 2α, b = cos 2β + i sin 2β and c = cos 2γ + i sin 2γ, prove that

Solution:

Question 6.
Solve: x⁴ + 4 = 0
Solution:

Question 7.
If x = a + b, y = aω + bω², z = aω² + bω, show that
(i) xyz = a³ + b³
(ii) x³ + y³ + z³ = 3(a³ + b³)
Solution:
(i) x = a + A; y = aω + bω², z = aω² + bω
Now xyz = (a + b) (aω + bω²) (abω² + bω) = (a+A) [aω³ + abω² + abω + b²ω³]
= (a + b) (a² – ab + b²) = a³ + b³
xyz = a³ + b³

(ii) x = a + b, y = aω + bω², z = aω² + bω
x + y + z = (a + aω + aω²) + (b + bω² + bω)
= a (1 + ω + ω²) + b (1 + ω + ω²) = a(0) + b(0) = 0
Now x + y + z = 0 ⇒ x³ + y³ + z³ = 3xyz
Here xyz = a³ + b³
∴ x³ + y³ + z³ = 3 (a³ + b³)

You can Download Samacheer Kalvi 12th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 12 Discrete Mathematics Ex 12.3

Choose the correct or the most suitable answer from the given four alternatives.

Question 1.
A binary operation on a set S is a function from …….
(a) S ➝ S
(b) (S × S) ➝ S
(c)S ➝ (S × S)
(d) (S × S) ➝ (S × S)
Solution:
(b) (S × S) ➝ S

Question 2.
Subtraction is not a binary operation in
(a) R
(b) Z
(c) N
(d) Q
Solution:
(c) N
Hint:
For example 2, 5 ∈ N but 2 – 5 = 3 ∉ N

Question 3.
Which one of the following is a binary operation on N ?
(a) Subtraction
(b) Multiplication
(c) Division
(c) All the above
Solution:
(b) Multiplication

Question 4.
In the set R of real numbers ‘*’ is defined as follows. Which one of the following is not a binary operation on R ?
(a) a * b = min (a.b)
(b) a * b = max (a, b)
(c) a * b = a
(d) a * b = a^b
Solution:
(d) a * b = a^b
Hint:
Since -2, 1/2 ∈ R , but (-2)^1/2 ∉ R.

Question 5.
The operation * defined by a * b = \(\frac{a b}{7}\) is not a binary operation on ……….
(a) Q⁺
(b) Z
(c) R
(c) C
Solution:
(b) Z
Hint:
Since 3, 5 ∈ Z, but \(\frac{3 \times 5}{7} \notin\) Z.

Question 6.



Solution:
(b) y = \(\frac{-2}{3}\)

Question 7.
If a * b = \(\sqrt{a^{2}+b^{2}}\) on the real numbers then * is ……..
(a) commutative but not associative
(b) associative but not commutative
(c) both commutative and associative
(d) neither commutative nor associative
Let, a, b ∈ R

(1) = (2) = * is associative
So * is both commutative and associative
Solution:
(c) both commutative and associative

Question 8.
Which one of the following statements has the truth value T ?
(a) sin x is an even function.
(b) Every square matrix is non-singular
(c) The product of complex number and its conjugate is purely imaginary
(d) \(\sqrt{5}\) is an irrational number
Solution:
(d) \(\sqrt{5}\) is an irrational number

Question 9.
Which one of the following statements has truth value F ?
(a) Chennai is in India or \(\sqrt{2}\) is an integer
(b) Chennai is in India or \(\sqrt{2}\) is an irrational number
(c) Chennai is in China or \(\sqrt{2}\) is an integer
(d) Chennai is in China or \(\sqrt{2}\) is an irrational number
Solution:
(c) Chennai is in China or \(\sqrt{2}\) is an integer

Question 10.
If a compound statement involves 3 simple statements, then the number of rows in the truth table is ……….
(a) 9
(b) 8
(c) 6
(d) 3
Solution:
(b) 8
Hint:
(i.e.) 2³ = 8

Question 11.
Which one is the inverse of the statement

Solution:
(a) \((\neg p \wedge \neg q) \rightarrow(\neg p \vee \neg q)\)

Question 12.
Which one is the contrapositive of the statement \((p \vee q) \rightarrow r\)?

Solution:
(a) \(\neg r \rightarrow(\neg p \wedge \neg q)\)

Question 13.
The truth table for \((p \wedge q) \vee \neg q\) is given below

Which one of the following is true?

Hint: The truth table for \((p \wedge q) \vee \neg q\)

Solution:
(3) T T F T

Question 14.
In the last column of the truth table for \(\neg(p \vee \neg q)\) the number of final outcomes of the truth value ‘F’ are
(a) 1
(b) 2
(c) 3
(d) 4
Hint:
The truth table for \(\neg(p \vee \neg q)\)

Solution:
(c) 3

Question 15.
Which one of the following is incorrect? For any two propositions p and q, we have …….

Solution:
(c) \(\neg(p \vee q) \equiv \neg p \vee \neg q\)

Question 16.

Which of the following is correct for the truth \((p \wedge q) \rightarrow \neg p\) ?

Hint:

Solution:
(2) F T T T

Question 17.


Solution:
(d) \(\neg(p \wedge q) \wedge | p \wedge(p \vee \neg r)]\)

Question 18.
The proposition \(p \wedge(\neg p \vee q)]\) is ……..
(a) a tautology
(b) a contradiction
(c) logically equivalent to \(p \wedge q\)
(d) logically equivalent to \(p \vee q\)
Solution:
(c) logically equivalent to \(p \wedge q\)

Question 19.
Determine the truth value of each of the following statements:
(a) 4 + 2 = 5 and 6+ 3 = 9
(b) 3 + 2 = 5 and 6 + 1 = 7
(c) 4 + 5 = 9 and 1 + 2 = 4
(d) 3 + 2 = 5 and 4 + 7 = 11

Solution:
(1) F T F T

Question 20.
Which one of the following is not true?
(a) Negation of a negation of a statement is the statement itself.
(b) If the last column of the truth table contains only T then it is a tautology.
(c) If the last column of its truth table contains only F then it is a contradiction
(d) If p and q are any two statements then p ⟷ q is a tautology.
Solution:
(d) If p and q are any two statements then p ⟷ q is a tautology.

Samacheer Kalvi 12th Maths Solutions Chapter 12 Discrete Mathematics Ex 12.3 Additional Problems

Choose the correct or the most suitable answer from the given four alternatives.

Question 1.
Which of the following are statements?
(i) May God bless you
(ii) Rose is a flower
(iii) milk is white
(iv) 1 is a prime number
(a) (i), (ii), (iii)
(b) (i), (ii), (iv)
(c) (i), (iii), (iv)
(d) (ii), (iii), (iv)
Hint:
Sentence (ii), (iii) and (iv) are statements
(ii) Rose is a flower – True
(iii) Milk is white – True
(iv) 1 is a prime number
∴ (ii), (iii), (iv) are statements
(i) May god bless you. This statement can not be assigned True or False.
∴ (i) is not a statements
Solution:
(d) (ii), (iii), (iv)

Question 2.
If a compound statement is made up of the three simple statements, then the number of rows in the truth table is …….
(a) 8
(b) 6
(c) 4
(d) 2
Hint:
The number of rows in truth table = 2ⁿ = 2³ = 8
Solution:
(a) 8

Question 3.
If p is T and q is F, then which of the following have the truth value T? ………

(a) (i), (ii), (iii)
(b) (i), (ii), (iv)
(c) (i), (iii), (iv)
(d) (ii), (iii), (iv)
Hint:
p is T then ~p is F
q is F then ~ q is T

Solution:
(c) (i), (iii), (iv)

Question 4.
The number of rows in the truth offimg6 is ……..
(a) 2
(b) 4
(c) 6
(d) 8
Hint:
Number of simple statements given is 2. i.e., p and q.
Number of rows in the truth table of \(\sim[p \wedge(\sim q)]\) = 2² = 4
Solution:
(b) 4

Question 5.
The conditional statement p ➝ q is equivalent to ……..


The truth table for p ➝ q and \((\sim p \vee q)\) having the last column identical.
∴ p ➝ q is equivalent to \((\sim p \vee q)\)
Solution:
(3) \(\sim p \vee q\)

Question 6.
Which of the following is a tautology?

Solution:
(c) \(\boldsymbol{p} \vee \sim \boldsymbol{p}\)
Hint:
A statement is said to be a tautology if the last column of its truth table contains only T.

Question 7.
In the set of integers with operation * defined by a * b = a + b – ab, the value of 3 * (4 * 5) is ……..
(a) 25
(b) 15
(c) 10
(d) 5
Hint:
a * b = a + b – ab
3 * (4 * 5) = 3 * (4 + 5 – 4(5))
= 3 * (9 – 20)
= 3 * (-11)
= 3 + (-11) – 3(-11)
= 3 – 11 + 33
= -8 + 33 = 25
Solution:
(a) 25

Question 8.
In the multiplicative group of cube root of unity, the order of \(\omega^{2}\) is ……
(a) 4
(b) 3
(c) 2
(d) 1
Solution:
(b) 3
Hint:

Question 10.

(a) 5
(b) 5\(\sqrt{2}\)
(c) 25
(d) 50
Hint:

Solution:
(b) 5\(\sqrt{2}\)

Question 11.
The order of -i in the multiplicative group of 4^th roots of unity is ……..
(a) 4
(b) 3
(c) 2
(d) 1
Hint:
The roots of fourth roots of unity are 1, -1, i, -i
The identity element is 1
(-i)⁴ = i⁴ = 1
Order of (-i) = 4.
Solution:
(a) 4

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Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 2 Complex Numbers Ex 2.7

Question 1.
Write in polar form of the following complex numbers.
(i) 2 + i2√3
(ii) 3 – i√3
(iii) -2 – i2
(iv) \(\frac{i-1}{\cos \frac{\pi}{3}+i \sin \frac{\pi}{3}}\)
Solution:

Question 2.
Find the rectangular form of the following complex numbers.

Solution:

Question 3.

Solution:
(i) (x₁ + iy₁) (x₂ + iy₂) (x₃ + iy₃) …….. (x_n + iy_n) = a + ib …… (1)
Taking modulus on both sides,
|(x₁ + iy₁) (x₂ + iy₂) (x₃ + iy₃) …….. (x_n + iy_n)| = |a + ib|
|x₁ + iy₁| |x₂ + iy₂| |x₃ + iy₃| ….. |x_n + iy_n| = |a + ib|

Question 4.
If \(\frac{1+z}{1-z}\) = cos 2θ + i sin 2θ, show that z = i tan θ.
Solution:

Question 5.
If cos α + cos β + cos γ = sin α + sin β + sin γ = 0, then show that
(i) cos 3α + cos 3β + cos 3γ = 3 cos (α + β + γ)
(ii) sin 3α + sin 3β + sin 3γ = 3 sin (α + β + γ)
Solution:
Let a = cos α + i sin α = e^iα
b = cos β + i sin β = e^iβ
c = cos γ + i sin γ = e^iγ
a + b + c = (cos α + cos β + cos γ) + i (sin α + sin β + sin γ)
⇒ a + b + c = 0 + i 0
⇒ a + b + c = 0
If a + b + c = 0 then a³ + b³ + c³ = 3abc

(cos 3α + i sin 3α + cos 3β + i sin 3β + cos 3γ + i sin 3γ) = 3 [cos (α + β + γ) + i sin (α + β + γ)]
(cos 3α + cos 3β + cos 3γ) + i (sin 3α + sin 3β + sin 3γ) = 3 cos (α + β + γ) + i 3sin(α + β + γ)
Equating real and Imaginary parts
(i) cos 3α + cos 3β + cos 3γ = 3 cos (α + β + γ)
(ii) sin 3α + sin 3β + sin 3γ = 3 sin (α + β + γ)

Question 6.
If z = x + iy and arg \(\left(\frac{z-i}{z+2}\right)=\frac{\pi}{4}\), then show that x² + y² + 3x – 3y + 2 = 0.
Solution:
arg \(\left(\frac{z-i}{z+2}\right)=\frac{\pi}{4}\)
We have arg (\(\frac{z_{1}}{z_{2}}\)) = arg(z₁) – arg(z₂)
arg (z – i) – arg (z + 2) = \(\frac{\pi}{4}\)
Let z = x + iy
arg (x + iy – i) – arg (x + iy + 2) = \(\frac{\pi}{4}\)
arg(x + i(y – 1)) – arg(x + 2 + iy) = \(\frac{\pi}{4}\)

2y – x – 2 = x² + y² + 2x – y
x² + y² + 3x – 3y + 2 = 0
Hence proved.

Samacheer Kalvi 12th Maths Solutions Chapter 2 Complex Numbers Ex 2.7 Additional Problems

Question 1.
Write the following complex numbers in the polar form:

Solution:

Question 2.
Find the modulus and principal argument of (1 + i) and hence express it in the polar form.
Solution:

Question 3.
Express the following complex numbers in the polar form.

Solution:

Question 4.
Express the following complex numbers in the polar form: \(2+2 \sqrt{3} i\)
Solution:

Question 5.
Express the following complex numbers in the polar form: \(-1+i \sqrt{3}\)
Solution:

Question 6.
Express the following complex numbers in the polar form: -1 – i
Solution:

Question 7.
Express the following complex numbers in the polar form: 1 – i
Solution:

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Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 2 Complex Numbers Ex 2.6

Question 1.
If z = x + iy is a complex number such that \(\left|\frac{z-4 i}{z+4 i}\right|=1\). show that the locus of z is real axis.
Solution:
\(\left|\frac{z-4 i}{z+4 i}\right|=1\)
⇒ |z – 4i| = |z + 4i|
let z = x + iy
⇒ |x + iy – 4i| = |x + iy + 4i|
⇒ |x + i(y – 4)| = |x +(y + 4)|
⇒ \(\sqrt{x^{2}+(y-4)^{2}}=\sqrt{x^{2}+(y+4)^{2}}\)
Squaring on both sides, we get
x² + y² – 8y + 16 = x² + y² + 16 + 8y
⇒ -16y = 0
⇒ y = 0 in two equation of real axis.

Question 2.
If z = x + iy is a complex number such that Im \(\left(\frac{2 z+1}{i z+1}\right)=0\) show that the locus of z is 2x² + 2y² + x – 2y = 0.
Solution:
Let z = x + iy

2x² + 2y² + x – 2y = 0.
Hence proved.

Question 3.
Obtain the Cartesian form of the locus of z = x + iy in each of the following cases:
(i) [Re(iz)]² = 3
(ii) Im[(1 – i)z + 1] = 0
(iii) |z + i| = |z – 1|
(iv) \(\bar{z}=z^{-1}\)
Solution:
(i) z = x + iy
[Re(iz)]² = 3
⇒ [Re[i(x + iy]]² = 3
⇒ [Re(ix – y)]² = 3
⇒ (-y)² = 3
⇒ y² = 3

(ii) Im[(1 – i)z + 1] = 0
⇒ Im [(1 – i)(z + iy) + 1] = 0
⇒ Im[x + iy – ix + y + 1] = 0
⇒ Im[(x + y + 1) + i(y – x)] = 0
Considering only the imaginary part
y – x = 0 ⇒ x = y

(iii) |z + i| = |z – 1|
⇒ |x + iy + i| = | x + iy – 1|
⇒ |x + i(y + 1)| = |(x – 1) + iy|
Squaring on both sides
|x + i(y + 1)|² = |(x – 1) + iy|²
⇒ x² + (y + 1)² = (x – 1)² + y²
⇒ x² + y² + 2y + 1 = x² – 2x + 1 + y²
⇒ 2y + 2x = 0
⇒ x + y = 0

(iv) \(\bar{z}=z^{-1}\)
⇒ \(\bar{z}=\frac{1}{z}\)
⇒ \(z \bar{z}=1\)
⇒ |z|² = 1
⇒ |x + iy|² = 1
⇒ x² + y² = 1

Question 4.
Show that the following equations represent a circle, and, find its centre and radius.
(i) |z – 2 – i| = 3
(ii) |2z + 2 – 4i| = 2
(iii) |3z – 6 + 12i| = 8
Solution:
(i) Let z = x + iy
|z – 2 – i| = 3
⇒ |x + iy – 2 – i| = 3
⇒ |(x – 2) + i(y – 1)| = 3
⇒ \(\sqrt{(x-2)^{2}+(y-1)^{2}}=3\)
Squaring on both sides
(x – 2)² + (y – 1)² = 9
⇒ x² – 4x + 4 + y² – 2y + 1 – 9 = 0
⇒ x² + y² – 4x – 2y – 4 = 0 represents a circle
2g = -4 ⇒ g = -2
2f = -2 ⇒ f = -1
c = -4
(a) Centre (-g, -f) = (2, 1) = 2 + i
(b) Radius = \(\sqrt{g^{2}+f^{2}-c}=\sqrt{4+1+4}=3\)
Aliter: |z – (2 + i)| = 3
Centre = 2 + i
radius = 3

(ii) |2(x + iy) + 2 – 4i| = 2
⇒ |2x + i2y + 2 – 4i| =2
⇒ |(2x + 2) + i(2y – 4)| = 2
⇒ |2(x + 1) + 2i(y – 2)| = 2
⇒ |(x + 1) + i(y – 2)| = 1
⇒ \(\sqrt{(x+1)^{2}(y-2)^{2}}=1\)
Squaring on both sides,
x² + 2x + 1 + y² + 4 – 4y – 1 = 0
⇒ x² + y² + 2x – 4y + 4 = 0 represents a circle
2g = 2 ⇒ g = 1
2f = -4 ⇒ f = -2
c = 4
(a) Centre (-g, -f) = (-1, 2) = -1 + 2i
(b) Radius = \(\sqrt{g^{2}+f^{2}-c}=\sqrt{1+4-4}=1\)
Aliter: 2|(z + 1 – 2i)| = 2
|z – (-1 + 2i)| = 1
Centre = -1 + 2i
radius = 1

(iii) |3(x + iy) – 6 + 12i| = 8
⇒ |3x + i3y – 6 + 12i| = 8
⇒ |3(x – 2) + i3 (y + 4)| = 8
⇒ 3|(x – 2) + i (y + 4)| = 8
⇒ \(3 \sqrt{(x-2)^{2}+(y+4)^{2}}=8\)
Squaring on both sides,
9[(x – 2)² + (y + 4)²] = 64
⇒ x² – 4x + 4 + y² + 8y + 16 = \(\frac{64}{9}\)
⇒ x² + y² – 4x + 8y + 20 – \(\frac{64}{9}\) = 0
x² + y² – 4x + 8y + \(\frac{116}{9}\) = 0 represents a circle.
2g = -4 ⇒ g = -2
2f = 8 ⇒ f = 4
c = \(\frac{116}{9}\)
(a) Centre (-g, -f) = (2, -4) = 2 – 4i
(b) Radius = \(=\sqrt{g^{2}+f^{2}-c}=\sqrt{4+16-\frac{116}{9}}=\sqrt{\frac{180-116}{9}}=\frac{8}{3}\)
Aliter:
|z – 2 + 4i| = \(\frac{8}{3}\)
⇒ |z – (2 – 4i)| = \(\frac{8}{3}\)
Centre = 2 – 4i, Radius = \(\frac{8}{3}\)

Question 5.
Obtain the Cartesian equation for the locus of z = x + iy in each of the following cases.
(i) |z – 4| = 16
(ii) |z – 4|² – |z – 1|² = 16
Solution:
(i) z = x + iy
|z – 4| = 16
⇒ |x + iy – 4| = 16
⇒ |(x – 4) + iy| = 16
⇒ \(\sqrt{(x-4)^{2}+y^{2}}=16\)
Squaring on both sides
(x – 4)² + y² = 256
⇒ x² – 8x + 16 + y² – 256 = 0
⇒ x² + y² – 8x – 240 = 0 represents the equation of circle

(ii) |x + iy – 4|² – |x + iy – 1|² = 16
⇒ |(x – 4) + iy|² – |(x – 1) + iy|² = 16
⇒ [(x – 4)² + y²] – [(x – 1)² + y²] = 16
⇒ (x² – 8x + 16 + y²) – (x² – 2x + 1 + y²) = 16
⇒ x² + y² – 8x + 16 – x² + 2x – 1 – y² = 16
⇒ -6x + 15 = 16
⇒ 6x + 1 = 0

Samacheer Kalvi 12th Maths Solutions Chapter 2 Complex Numbers Ex 2.6 Additional Problems

Question 1.
If the imaginary part of is -2, then show that the locus of the point representing z in the argand plane is a straight line.
Solution:
Let z = x + iy. Then,


Hence, the locus of z is a straight line

Question 2.
If the real part of is 4, then show that locus of the point representing z in the complex plane is a circle.
Solution:

It is given that the real part of is 4.

Question 3.

Solution:

Question 4.
If arg (z – 1) = \(\frac{\pi}{6}\) and arg (z + 1) = 2 \(\frac{\pi}{3}\) , then prove that |z| = 1.
Solution:

Question 5.
P represents the variable complex number z. Find the locus of P, if
Solution:

Question 6.
P represents the variable complex number z. Find the locus of P, if
Solution:
Let z = x + iy

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Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 12 Discrete Mathematics Ex 12.2

Question 1.
Let p : Jupiter is a planet and q : India is an island be any two simple statements. Give verbal sentence describing each of the following statements.

Solution:
(i) \(\neg p\) : Jupiter is not a planet
(ii) \(p \wedge \neg q\) : Jupiter is not a planet and India is not an island
(iii) \(\neg p \vee q\) : Jupiter is not a planet or India is an island.
(iv) \(p \rightarrow \neg q\) : If Jupiter is a planet then India is not an island
(v) \(p \leftrightarrow q\) : If Jupiter is a planet if and only if India is an island

Question 2.
Write each of the following sentences in symbolic form using statement variables p and q.
(i) 19 is not a prime number and all the angles of a triangle are equal.
(ii) 19 is a prime number or all the angles of a triangle are not equal
(iii) 19 is a prime number and all the angles of a triangle are equal
(iv) 19 is not a prime number
Solution:
p : 19 is a prime number
q : All the angles of a triangle are equal

Question 3.
Determine the truth value of each of the following statements
(i) If 6 + 2 = 5 , then the milk is white.
(ii) China is in Europe or \(\sqrt{3}\) is an integer
(iii) It is not true that 5 + 5 = 9 or Earth is a planet
(iv) 11 is a prime number and all the sides of a rectangle are equal
Solution:

Question 4.
Which one of the following sentences is a proposition?
(i) 4 + 7 = 12
(ii) What are you doing?
(iii) 3ⁿ ≤ 81, n ∈ N
(iv) Peacock is our national bird
(v) How tall this mountain is!
Solution:
(i) is a proposition
(ii) not a proposition
(iii) is a proposition
(iv) is a proposition
(v) not a proposition

Question 5.
Write the converse, inverse, and contrapositive of each of the following implication.
(i) If x and y are numbers such that x = y, then x² = y²
(ii) If a quadrilateral is a square then it is a rectangle
Solution:
(i) Converse: If x and y are numbers such that x² = y² then x = y.
Inverse: If x and y are numbers such that x ≠ y then x² ≠ y².
Contrapositive : If x and v are numbers such that x² ≠ y² then x ≠ y.

(ii) Converse: If a quadrilateral is a rectangle then it is a square.
Inverse: If a quadrilateral is not a square then it is not a rectangle.
Contrapositive : If a quadrilateral is not a rectangle then it is not a square.

Question 6.
Construct the truth table for the following statements.

Solution:

Question 7.
Verify whether the following compound propositions are tautologies or contradictions or contingency

Solution:

In the above Truth table the last column entries are ‘F’. So the given propositions is a contradiction.

In the above truth table the last column entries are ‘T’. So the given propositions is a tautology.

In the above truth table the entries in the last column are a combination of’ T ‘ and ‘ F ‘. So the given statement is neither propositions is neither tautology nor a contradiction. It is a contingency.

The last column entires are ‘T’. So the given proposition is a tautology.

Question 8.

Solution:

Question 9.

Solution:

The entries in the column corresponding to q ➝ p and \(\neg p \rightarrow \neg q\) are identical and hence they are equivalent.

Question 10.
Show that p ➝ q and q ➝ p are not equivalent
Solution:

The entries in the column corresponding to p ➝ q and q ➝ p are not identical, hence they are not equivalent.

Question 11.

Solution:

Question 12.
Check whether the statement p ➝ (q ➝ p) is a tautology or a contradiction without using the truth table.
Solution:

Question 13.
Using truth table check whether the statements are logically equivalent.
Solution:

Question 14.
without using truth table
Solution:

Question 15.

Solution:

The entries in the column corresponding to are identical.
Hence they are equivalent.

Samacheer Kalvi 12th Maths Solutions Chapter 12 Discrete Mathematics Ex 12.2 Additional Problems

Question 1.
Show that is a tautology.
Solution:

Question 2.
Show that is a contradiction.
Solution:

Question 3.
Use the truth table to determine whether the statement is a tautology.
Solution:

The last column contains only T. ∴ The given statement is a tautology.

Question 4.
Show that
Solution:
(i) Truth table for p \(\leftrightarrow\) q

Question 5.
Show that .
Solution:


The last columns of statements (i) and (ii) are identical.

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Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 2 Complex Numbers Ex 2.5

Question 1.
Find the modulus of the following complex numbers.
(i) \(\frac{2 i}{3+4 i}\)
(ii) \(\frac{2-i}{1+i}+\frac{1-2 i}{1-i}\)
(iii) (1 – i)¹⁰
(iv) 2i(3 – 4i) (4 – 3i)
Solution:

(iii) |(1 – i)¹⁰| = (|1 – i|)¹⁰
= \((\sqrt{1+1})^{10}=(\sqrt{2})^{10}=2^{5}=32\)
(iv) |2i(3 – 4i) (4 – 3i)|
= |2i| |3 – 4i| |4 – 3i|
= \(2 \sqrt{9+16} \sqrt{16+9}\)
= 2 × 5 × 5
= 50

Question 2.
For any two complex numbers z₁ and z₂, such that |z₁| = |z₂| = 1 and z₁ z₂ ≠ -1, then show that \(\frac{z_{1}+z_{2}}{1+z_{1} z_{2}}\) is a real number.
Solution:
|z₁|² = 1
⇒ \(z_{1} \bar{z}_{1}=1\)
⇒ \(z_{1}=\frac{1}{\bar{z}_{1}}\)
Similarly \(z_{2}=\frac{1}{\bar{z}_{2}}\)

Question 3.
Which one of the points 10 – 8i, 11 + 6i is closest to 1 + i.
Solution:
A (1 + i), B (10 – 8i), C (11 + 6i)
|AB| = |(10 – 8i) – (1 + i)|
= |10 – 8i – 1 – i|
= |9 – 9i|
= \(\sqrt{81+81}\)
= \(\sqrt{162}\)
= 9(1.414)
= 12.726
CA = |(11 + 6i) – (1 + i)|
= |11 + 6i – 1 – i|
= |10 + 5i|
= \(\sqrt{100+25}\)
= \(\sqrt{125}\)
C (11 + 6i) is closest to the point A (1 + i)

Question 4.
If |z| = 3, show that 7 ≤ |z + 6 – 8i| ≤ 13.
Solution:
|z| = 3, To find the lower bound and upper bound we have
||z₁| – |z₂|| ≤ |z₁ + z₂| ≤ |z₁| + |z₂|
||z| – |6 – 8i|| ≤ |z + 6 – 8i| ≤ |z| + |6 – 8i|
|3 – \(\sqrt{36+64}\)| ≤ |z + 6 – 8i| ≤ 3 + \(\sqrt{36+64}\)
|3 – 10| ≤ |z + 6 – 8i| ≤ 3 + 10
7 ≤ |z + 6 – 8i| ≤ 13

Question 5.
If |z| = 1, show that 2 ≤ |z² – 3| ≤ 4.
Solution:
|z| = 1 ⇒ |z|² = 1
||z₁| – |z₂|| ≤ |z₁ + z₂| ≤ |z₁| + |z₂|
||z|² – |-3|| ≤ |z² – 3| ≤ |z|² + |-3|
|1 – 3| ≤ |z² – 3| ≤ 1 + 3
2 ≤ |z² – 3| ≤ 4

Question 6.
If \(\left|z-\frac{2}{z}\right|\) = 2, show that the greatest and least value of |z| are √3 + 1 and √3 – 1 respectively.
Solution:
\(\left|z-\frac{2}{z}\right|\) = 2
We know that

The minimum value of |z| is |1 – √3| = √3 – 1
The greatest value of |z| is √3 + 1

Question 7.
If z₁, z₂ and z₃ are three complex numbers such that |z₁| = 1, |z₂| = 2, |z₃| = 3 and |z₁ + z₂ + z₃| = 1, show that |9z₁ z₂ + 4z₁ z₃ + z₂ z₃| = 6.
Solution:

Question 8.
If the area of the triangle formed by the vertices z, iz, and z + iz is 50 square units, find the value of |z|.
Solution:
The given vertices are z, iz, z + iz ⇒ z, iz are ⊥r to each other.
Area of triangle = \(\frac { 1 }{ 2 }\) bh = 50
⇒ \(\frac { 1 }{ 2 }\) |z| |iz| = 50
⇒ \(\frac { 1 }{ 2 }\) |z| |z| = 50
⇒ |z|² = 100
⇒ |z| = 10

Question 9.
Show that the equation z³ + 2\(\bar{z}\) = 0 has five solutions.
Solution:
Given that z³ + 2\(\bar{z}\) = 0
z³ = -2 \(\bar{z}\) ……. (1)
Taking modulus on both sides,

z has four non-zero solution.
Hence including zero solution. There are five solutions.

Question 10.
Find the square roots of
(i) 4 + 3i
(ii) -6 + 8i
(iii) -5 – 12i
Solution:
(i) z = 4 + 3i
|z| = |4 + 3i| = \(\sqrt{16+9}\) = 5

Samacheer Kalvi 12th Maths Solutions Chapter 2 Complex Numbers Ex 2.5 Additional Problems

Question 1.
Find the modulus and argument of the following complex numbers and convert them in polar form.

Solution:
(i)

Question 2.
Find the square roots of – 15 – 8i
Solution:

On solving (i) and (iii), we get
x² = 1 and y² = 16 => x = ± 1 and y = ±4 From (ii), we observe that 2xy is negative. So, x and y are of opposite signs.

Question 3.
Find the square roots of i.
Solution:

From (ii) we observe that we find that 2xy is positive. So, x and y are of same sign.

Question 4.
Find the modulus or the absolute value of
Solution:

Question 5.
Find the modulus and argument of the following complex numbers:

Solution:

Question 6.
Show that the points representing the complex numbers 7 + 9i, – 3 + 7i, 3 + 3i form a right angled triangle on the Argand diagram.
Solution:
Let A, B and C represent the complex numbers
7 + 9i, – 3 + 7i and 3 + 3i in the Argand diagram respectively.


Hence ∆ABC is a right angled isosceles triangle.

Question 7.
Find the square root of (- 7 + 24i).
Solution:

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Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 12 Discrete Mathematics Ex 12.1

Question 1.
Determine whether * is a binary operation on the sets given below
(i) a * b = a.|b| on R ,
(ii) a * b = min (a, b) on A = {1, 2, 3, 4, 5}
(iii) (a * b) = \(a \sqrt{b}\) is binary on R.
Solution:
(i) Yes.
Reason: a, b ∈ R. So, |b| ∈ R when b ∈ R
Now multiplication is binary on R
So a|b| ∈ R when a,be R.
(Le.) a * b ∈ R.
* is a binary operation on R.

(ii) Yes.
Reason: a, b ∈ R and minimum of (a, b) is either a or b but a, b ∈ R.
So, min (a, b) ∈ R.
(Le.) a * b ∈ R.
* is a binary operation on R.

(iii) a* b = \(a \sqrt{b}\) where a, b ∈ R.
No. * is not a binary operation on R.
Reason: a, b ∈ R.
⇒ b can be -ve number also and square root of a negative number is not real.
So \(\sqrt{b}\) ∉ R even when b ∈ R.
So \(\sqrt{b}\) ∉ R. ie., a * b ∉ R.
* is not a binary operation on R.

Question 2.

Solution:
No. * is not a binary operation on Z.
Reason: Since m, n ∈ Z.
So, m, n can be negative also.
Now, if n is negative (Le.) say n = -k where k is +ve.

Similarly, when m is negative then n^m ∉ Z.
∴ m * n ∉ Z. ⇒ * is not a binary operation on Z.

Question 3.
Let * be defined on R by (a * b) = a + b + ab – 1 .Is * binary on R ? If so, find \(3 *\left(\frac{-7}{15}\right)\)
Solution:
a * b = a + b + ab – 7.
Now when a, b ∈ R, then ab ∈ R also a + b ∈ R.
So, a + b + ab ∈ R.
We know – 7 ∈ R.
So, a + b + ab – 7 ∈ R.
(ie.) a * b ∈ R.
So, * is a binary operation on R.

Question 4.
Let A= {a+ \(\sqrt{5}\)b: a, b ∈ Z}. Check whether the usual multiplication is a binary operation on A.
Solution:
Let A = a + \(\sqrt{5}\) b and B = c + \(\sqrt{5}\)d, where a, b, c, d ∈ M.
Now A * B ={a + \(\sqrt{5}\)b)(c + \(\sqrt{5}\)d)
= ac + \(\sqrt{5}\)ad + \(\sqrt{5}\)bc + \(\sqrt{5}\)b\(\sqrt{5}\)d
= (ac + 5bd) + \(\sqrt{5}\)(ad+ bc) ∈ A
Where a, b, c, d ∈ Z
So * is a binary operation.

Question 5.
(i) Define an operation * on Q as follows: a * b = \(\left(\frac{a+b}{2}\right)\); a, b ∈ Q. Examine the closure,
commutative, and associative properties satisfied by * on Q.
(ii) Define an operation * on Q as follows: a*b = \(\left(\frac{a+b}{2}\right)\); a, b ∈ Q. Examine the existence of identity and the existence of inverse for the operation * on Q.
Solution:
(i) 1. Closure property:
Let a,b ∈ Q.

So, closure property is satisfied.

2. Commutative property:
Let a, b ∈ Q.

(1) = (2) ⇒ Now a * b = b * a
⇒ Commutative property is satisfied.

3. Associative property:
Let a,b,c G Q. ^
To prove associative property we have to prove that a * (b * c) = (a * b) * c
LHS: a * (b * c)


(i.e.) the identity Clement e = a which is not possible.
So, the identity element does not exist and so inverse does not exist.

Question 6.
Fill In the following table so that the binary operation * on A = {a, b, c} is commutative.

Solution:
Given that the binary operation * is Commutative.
To find a * b :
a * b = b * a (∵ * is a Commutative)
Here b * a = c. So a * b = c
To find a *c:

a * c = c * a (∵ * is a Commutative)
c * a = a. (Given)
So a * c = a
To find c * b:
c * b = b * c
Here b * c = a.
So c * b = a

Question 7.
Consider the binary operation * defined on the set A = [a, b, c, d] by the following table:

– Is it commutative and associative?
Solution:
From the table
b * c = b
c * b = d
So, the binary operation is not commutative.
To check whether the given operation is associative.
Let a, b, c ∈ A.
To prove the associative property we have to prove that a * (b * c) = (a * b) * c
From the table,
LHS: b * c = b
So, a * (b * c) = a * b = c ……. (1)
RHS: a * b = c
So, (a * b) * c = c * c = a …… (2)
(1) ≠ (2). So, a * (b * c) ≠ (a * b) * c
∴ The binary operation is not associative.

Question 8.

Solution:

Question 9.
and Let * be the matrix multiplication. Determine whether M is closed under * . If so, examine the commutative and associative properties satisfied by * on M .

and let * be the matrix multiplication. Determine whether M is closed under *. If so, examine the existence of identity, existence of inverse properties for the operation * on M.
Solution:



So, inverse property is satisfied.

Question 10.
(i) Let A be Q\{1). Define * on A by x * y = x + y – xy. Is * binary on A ? If so, examine the commutative and associative properties satisfied by * on A.
(ii) Let A be Q\{1}. Define *on A by x * y = x + y – xy. Is * binary on A ? If so, examine the existence of identity, existence of inverse properties for the operation * on A.
Solution:
(i) Let a,b ∈ A (i.e.) a ≠ ±1 , b ≠ 1
Now a * b = a + b – ab
If a + b – ab = 1 ⇒ a + b – ab – 1 = 0
(i.e.) a(1 – b) – 1(1 – b) = 0
(a – 1)(1 – b) = 0 ⇒ a = 1, b = 1
But a ≠ 1 , b ≠ 1
So (a – 1) (1 – 6) ≠ 1
(i.e.) a * b ∈ A. So * is a binary on A.

To verify the commutative property:

Let a, b ∈ A (i.e.) a ≠ 1 , b ≠ 1
Now a * b = a + b – ab
and b * a = b + a – ba
So a * b = b * a ⇒ * is commutative on A.

To verify the associative property:
Let a, b, c ∈ A (i.e.) a, b, c ≠ 1
To prove the associative property we have to prove that
a * (b * c) = (a * b) * c

LHS: b * c = b + c – bc = D(say)
So a * (b * c) = a * D = a + D – aD
= a + (b + c – bc) – a(b + c – bc)
= a + b + c – bc – ab – ac + abc
= a + b + c – ab – bc – ac + abc …… (1)

RHS: (a * b) = a + b – ab = K(say)
So (a * b) * c = K * c = K + c – Kc
= (a + b – ab) + c – (a + b – ab) c
= a + b – ab + c – ac – bc + abc
= a + b + c – ab – bc – ac + abc ….. (2)

(ii) To verify the identity property:
Let a ∈ A (a ≠ 1)
If possible let e ∈ A such that
a * e = e * a = a
To find e:
a * e = a
(i.e.) a + e – ae = a

So, e = (≠ 1) ∈ A
(i.e.) Identity property is verified.
To verify the inverse property:
Let a ∈ A (i.e. a ≠ 1)
If possible let a’ ∈ A such that
To find a’:
a * a’ = e
(i.e.) a + a’ – aa’ = 0
⇒ a'(1 – a) = – a

⇒ For every a ∈ A there is an inverse a’ ∈ A such that
a* a’ = a’ * a = e
⇒ Inverse property is verified.

Samacheer Kalvi 12th Maths Solutions Chapter 12 Discrete Mathematics Ex 12.1 Additional Problems

Question 1.
Show that the set G = {a + b\(\sqrt{2}\)/ a, b ∈ Q} is an infinite abelian group with respect to Binary operation addition. Satisfies closure, associative, identity and inverse properties.
Solution:
(i) Closure axiom :
Let x, y ∈ G. Then x = a + b\(\sqrt{2}\), y = c + d\(\sqrt{2}\); a, b, c, d ∈ Q.
x + y = (a + b\(\sqrt{2}\)) + (c + d\(\sqrt{2}\)) = (a + c) + (b + d) \(\sqrt{2}\) ∈ G,
Since (a + c) and (b + d) are rational numbers.
∴ G is closed with respect to addition.

(ii) Associative axiom : Since the elements of G are all real numbers, addition is associative.

(iii) Identity axiom : There exists 0 = 0 + 0 \(\sqrt{2}\) ∈ G
such that for all x = a + b\(\sqrt{2}\) ∈ G.
x + 0 = (a + b\(\sqrt{2}\) ) + (0 + 0\(\sqrt{2}\))
= a + b\(\sqrt{2}\) = x
Similarly, we have 0 + x = x. ∴ 0 is the identity element of G and satisfies the identity axiom.

(iv) Inverse axiom: For each x = a + b\(\sqrt{2}\) ∈ G,
there exists -x = (-a) + (-b) \(\sqrt{2}\) ∈ G
such that x + (-x) = (a + b\(\sqrt{2}\)) + ((-a) + (- b)\(\sqrt{2}\))
= (a + (-a)) + (b + (-b)) \(\sqrt{2}\) = 0
Similarly, we have (- x) + x = 0 .
∴ (- a) + (-b)\(\sqrt{2}\) is the inverse of a + b \(\sqrt{2}\) and satisfies the inverse axiom.

(v) Commutative axiom:
x + y = (a + c) + (b + d) \(\sqrt{2}\) = (c + a) + (d + b) \(\sqrt{2}\)
= (c + d\(\sqrt{2}\)) + (a + b\(\sqrt{2}\))
= y + x, for all x, y ∈ G.
∴ The commutative property is true.
∴ (G, +) is an abelian group. Since G is infinite, we see that (G, +) is an infinite abelian group.

Question 2.
Show that (Z₇ – { [0]}, .7) write to the binary operation multiplication modul07 satisfies closure, associative, identity and inverse properties.
Solution:
Let G = [[1], [2],… [6]]
The Cayley’s table is

From the table:
(i) all the elements of the composition table are the elements of G.
∴ The closure axiom is true.
(ii) multiplication modulo 7 is always associative.
(iii) the identity element is [1] ∈ G and satisfies the identity axiom.
(iv) the inverse of [1] is [1]; [2] is [4]; [3] is [5]; [4] is [2]; [5] is [3] and [6] is [6] and it satisfies the inverse axiom.

Question 3.
Show that the set G of all positive rationals with respect to composition * defined by ab
a* b = \(\frac{a b}{3}\) for all a, b ∈ G satisfies closure, associative, identity and inverse properties.
Solution:
Let G = Set of all positive rational number and * is defined by,

(i) Closure axiom: Let a, b ∈ G

∴ closure axiom is satisfied.

(ii) Associative axiom: Let a, b, c ∈ G.
To prove the associative property, we have to prove that

(1) = (2) ⇒ LHS = RHS i.e., associative axiom is satisfied.

(iii) Identity axiom: Let a ∈ G.
Let e be the identity element.
By the definition, a * e = a

e = 3 ∈ G ⇒ identity axiom is satisfied.

(iv) Inverse axiom: Let a ∈ G and a’ be the inverse of a * a’ = e = 3.

Question 4.
Show that the set G of all rational numbers except – 1 satisfies closure, associative, identity and inverse property with respect to the operation * given by a * b = a + b + ab for all a, b ∈ G
Solution:
G = [Q, -{-l}]
* is defined by a * b = a + b + ab
To prove G is an abelian group.

G₁: Closure axiom: Let a, b ∈ G.
i.e., a and b are rational numbers and a ≠ -1, b ≠ -1.
So, a * b = a + b + ab
If a + b + ab = – 1
⇒ a + b + ab + 1 = 0
i.e., (a + ab) + (b + 1) = 0
a (1 + b) + (b + 1) = 0
i.e., (a + 1)(1 + b) = 0
⇒ a = -1, b = -1
But a ≠ -1, b ≠ -1
⇒ a + b + ab ≠ -1
i.e., a + b + ab ∈ G ∀ a, b ∈ G
⇒ Closure axiom is verified.

G₂: Associative axiom: Let a, b, c ∈ G.
To prove G₂, we have to prove that,
a * {b * c) = (a * b) * c
LHS:
b * c = b + c + bc = D (say)
a * (b * c) = a * D = a + D + aD
= a + (b + c + bc) + a(b + c + bc)
= a + b + c + bc + ab + ac + abc
= a + b + c + ab + bc + ac + abc ……. (1)
RHS:
a * b = a + b + ab = E (say)
∴ (a * b) * c = E * c = E + c + Ec
= a + b + ab + c + (a + b + ab) c
= a + b + ab + c + ac + bc + abc ……. (2)
= a + b + c + ab +be+ ac + abc
(1) = (2) ⇒ Associative axiom is verified.

G₃: Identity axiom: Let a ∈ G. To prove G₃ we have to prove that there exists an element e ∈ G such that a * e = e * a = a.
To find e: a * e = a
i.e., a + e + ae = a
⇒ e(1 + a) = a – a = 0

So, e = 0 ∈ G ⇒ Identity axiom is verified.

G₄ : Inverse axiom: Let a ∈ G. To prove G₄, we have to prove that there exists an element a’ ∈ G such that a * a’ = a’ * a = e.
To find a’: a * a’ = e
i.e., a + a’ + aa’=: 0 {∵ e = 0}
⇒ a'(1 + a) = -a

Thus, inverse axiom is verified.

Question 5.
Show that the set { [1], [3], [4], [5], [9]} under multiplication modulo 11 satisfies closure, associative, identity and inverse properties.
Solution:
G = {[1], [3], [4], [5], [9]}
* is defined by multiplication modulo 11.
To prove G is an abelian group with respect to *
Since we are given a finite number of elements i.e., since the given set is finite, we can frame the multiplication table called Cayley’s table.
The Cayle’s table is as follows:

G₁: The elements in the above table are [1], [3], [4], [5] and [9] which are elements of G.
∴ closure axiom is verified.

G₂: Consider [3], [4], [5] which are elements of G.
{[3] * [4]} * [5] = [1] * [5] = [5] ……. (1)
[3] * {[4] * [5]} = [3] * [9] = [5] …… (2)
(1) = (2) ⇒ (a * b) * c = a * (b * c) i.e., associative axiom is verified.

G₃: The first row elements are the same as that of the given elements in the same order. ie., from the table, the identity element is [1] ∈ G. So identity axiom is verified.

G₅: From the table * is commutative i.e., the entries equidistant from the leading diagonal on either sides are equal ⇒ a * b = b * a

You can Download Samacheer Kalvi 12th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 2 Complex Numbers Ex 2.4

Question 1.
Write the following in the rectangular form:
(i) \(\overline{(5+9 i)+(2-4 i)}\)
(ii) \(\frac{10-5 i}{6+2 i}\)
(iii) \(\overline{3 i}+\frac{1}{2-i}\)
Solution:

Question 2.
If z = x + iy, find the following in rectangular form.
(i) Re(\(\frac{1}{z}\))
(ii) Re(i\(\bar{z}\))
(iii) Im(3z + 4\(\bar{z}\) – 4i)
Solution:
(i) Re(\(\frac{1}{z}\)) = Re(\(\frac{1}{x+i y} \times \frac{x-i y}{x-i y}\))
= Re(\(\frac{x-i y}{x^{2}+y^{2}}\))
= \(\frac{x}{x^{2}+y^{2}}\)
(ii) Re(i\(\bar{z}\)) = Re[i(\(\overline{x+i y}\))]
= Re(ix + y)
= y
(iii) Im(3z + 4\(\bar{z}\) – 4i)
= Im (3(x + iy) + 4(x – iy) – 4i)
= Im (3x + 3iy + 4x – 4iy – 4i)
= Im (3x + 4 + i (3y – 4y – 4)
= Im (3x + 4x + i(-y – 4))
= Im [7x + i(-y – 4)]
= -y – 4
= -(y + 4)

Question 3.
If z₁ = 2 – i and z₂ = -4 + 3i, find the inverse of z₁ z₂ and \(\frac{z_{1}}{z_{2}}\)
Solution:
z₁ = 2 – i, z₂ = -4 + 3i
(i) z₁ z₂ = (2 – i) (-4 + 3i)
= (-8 + 6i + 4i – 3 i²)
= (-8 + 10i + 3)
= (-5 + 10i)

Question 4.
The complex numbers u, v, and w are related by \(\frac{1}{u}=\frac{1}{v}+\frac{1}{w}\). If v = 3 – 4i and w = 4 + 3i, find u in rectangular form.
Solution:
v = 3 – 4i, w = 4 + 3i = i (3 – 4i)

Question 5.
Prove the following properties:
(i) z is real if and only if z = \(\bar{z}\)
(ii) Re(z) = \(\frac{z+\bar{z}}{2}\) and Im(z) = \(\frac{z-\bar{z}}{2 i}\)
Solution:
(i) z is real iff z = \(\bar{z}\)
Let z = x + iy
z = \(\bar{z}\)
⇒ x + iy = x – iy
⇒ 2iy = 0
⇒ y = 0
⇒ z is real.
z is real iff z = \(\bar{z}\)
(ii) \(\frac{z+\bar{z}}{2 i}=\frac{x+i y+x-i y}{2}=\frac{2 x}{2}=x\)
Real part of z = x
(iii) \(\frac{z-\bar{z}}{2 i}=\frac{(x+i y)-(x-i y)}{2 i}=\frac{x+i y-x+i y}{2 i}=\frac{2 i y}{2 i}=y\)
Im part of z = y.

Question 6.
Find the least value of the positive integer n for which (√3 + i)ⁿ
(i) real
(ii) purely imaginary
Solution:
(√3 + i)ⁿ
(√3 + i)²
= 3 – 1 + 2√3 i
= (2 + 2 √3 i)
(√3 + i)³ = (√3 + i)² (√3 + i)
= (2 + 2√3 i) (√3 + i)
= 2√3 + 2i + 6i – 2√3
(√3 + i) = 8i ⇒ purely Imaginary when n = 3
(√3 + i)⁴ = (√3 + i)³ (√3 + i)
= 8i (√3 + i)
= (-8 + 8√3 i)
(√3 + i)⁵ =(√3 + i)⁴ (√3 + i)
= (-8 + 8√3 i) (√3 + i)
= -8√3 – 8i + 24i – 8√3
= -16√3 + 16i
(√3 + i)⁶ = (√3 + i)⁵ (√3 + i)
= (√3 + i) (-16√3 + 16i)
= 16 (√3 + i) (-√3 + i)
= 16 (-3 + i√3 – i√3 – 1)
= -64 purely real when n = 6
Another Method:

Question 7.
Show that
(i) (2 + i√3)¹⁰ – (2 – i√3)¹⁰ is purely imaginary
(ii) \(\left(\frac{19-7 i}{9+i}\right)^{12}+\left(\frac{20-5 i}{7-6 i}\right)^{12}\)
Solution:
(i) (2 + i√3)¹⁰ – (2 – i√3)¹⁰

Samacheer Kalvi 12th Maths Solutions Chapter 2 Complex Numbers Ex 2.4 Additional Problems

Question 1.
Express the following in the standard form a + ib.

Solution:

Question 2.
Find the least positive integer n such that
Solution:

Question 3.
Find the real values of x and y for which the following equations are satisfied.

Solution:
(i) (1 – i)x + (1 + i)y = x – ix + y + iy
= (x + y) + i (y – x) = 1 – 3i (given)
So, equating their RP and IP we get,

Take real part, we get
i.e., 3x + (x – 2) + 6y – (1 – 3y) = 0
⇒ 3x + x – 2 + 6y – 1 + 3y = 0
4x + 9y = 3 …….. (1)
Take imaginary part, we get
3(x – 2) – x + 3 (1 – 3y) + 2y = 10
⇒ 3x – 6 – x + 3 – 9y + 2y = 10

Squaring on both sides, x² + 3x + 8 = 4 (x + 4)²
i.e., x² + 3x +8 = 4 (x² + 8x +16) ⇒ 4x² + 32x + 64 – x² – 3x – 8 = 0
3x² + 29x + 56 = 0
3x² + 21x + 8x + 56 = 0
(x + 7) (3x + 8) = 0