Category: Class 12

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 2 Complex Numbers Ex 2.8 Textbook Questions and Answers, Notes.

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

Solution:
L.H.S

Question 2.
Show that

Solution:

Question 3.

Solution:


Aliter method:

Question 4.
If 2cos α = x + \(\frac{1}{x}\) and 2 cos β = y + \(\frac{1}{x}\), show that
(i) \(\frac{x}{y}+\frac{y}{x}=2 \cos (\alpha-\beta)\)
Solution:
Given 2 cos α = x + \(\frac{1}{x}\) and cos β = y + \(\frac{1}{y}\)
simplifying x² – 2x cos α + 1 = 0

if x = cos α + i sin α, then \(\frac{1}{x}\) = cos α – i sin α
similarly y = cos β + i sin β and \(\frac{1}{y}\) = cos β – i sin β

Hence proved

(ii) xy = (cos α + i sin α)(cos β + i sin β )
Solution:
xy = (cos α + i sin α) (cos β + i sin β)
= cos (α + β) + i sin (α + β)
[∵ arg (z₁z₂) = arg z₁ + arg z₂
\(\frac{1}{xy}\) = cos (α + β) – i sin (α + β)
∴ xy – \(\frac{1}{xy}\) = cos (α + β) + i sin (α + β) – cos (α + β) + i sin (α + β)
= 2i sin (α + β)
Hence proved

(iii) \(\frac{x^{m}}{y^{n}}-\frac{y^{n}}{x^{m}}\) = 2 i sin (mα – nβ)
Solution:

Hence Proved

(iv) x^m yⁿ + \(\frac { 1 }{ x^m y^n }\) = 2 cos(mα – nβ)
Solution:
= (cos mα + sin mα) (cos nβ + i sin nβ)
= cos (mα + nβ) + i sin (mα + nβ)

Hence proved

Question 5.
Solve the equation z³ + 27 = 0.
Solution:
z³ + 27 = 0
z³ = – 27 = 27 (-1)
= 27 [cos(π + 2kπ) + i sin(π + 2kπ)], k ∈ z
∴ z = (27)^1/3[cos (2k + 1)π + i sin (2k+1)π]^1/3 k ∈ z

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:
Given ω ≠ 1 is a active root of unity
(z – 1)³ + 8 = 0
(z- 1)³ = -8
z – 1 = (-8)^1/3 (1)^1/3
= (-2) (1, ω, ω²)
z – 1 = (-2, -2ω, -2ω²)
= z – 1 = -2
z = -2 + 1 = -1
z – 1 = -2ω
z = 1 – 2ω
z – 1 = -2ω²
z = 1 – 2ω²

Question 7.

Solution:
\(\sum_{k=1}^{8}\) (cos \(\frac { 2kπ }{ 9 }\) + i sin \(\frac { 2kπ }{ 9 }\))
The sum all nth root of unity is
1 + ω + ω² + …….. + ω^n-1 = 0
From the given polar from , it is clear that the complex number is 1 + i0 (unity)

Question 8.
If ω ≠ 1 is a cube root of unity, show that
(i) (1 – ω + ω²)⁶ + (1 + ω – ω²)⁶ = 128.
Solution:
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
Solution:
(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
Hence proved.

Question 9.
If z = 2 – 2i, find the rotation of z by θ radians in the counterclockwise direction about the origin when
Solution:
Let 2 – 2i
Modules = |z| = \(\sqrt{2^2+2^2}\) = 2√2
Argument θ = tan^-1(\(\frac{-2}{2}\)) = tan^-1(-1) = –\(\frac{π}{4}\)
(i) when ‘z’ is rotated in the counter clockwise direction about the origin when θ = \(\frac{π}{3}\) i.,e argument of new position

(ii) θ = \(\frac{2π}{3}\)
Solution:

(iii) θ = \(\frac{3π}{3}\)
Solution:

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 2 Complex Numbers Ex 2.7 Textbook Questions and Answers, Notes.

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
Solution:
Let z + i2√3 = r (cos θ + i sin θ)
equating real and imaginary parts
r cos θ = 2 (+ve)
r sin θ = 2√3 (+ve)
r² cos² θ + r² sin² θ = (2)² + (2√3)²
r² = 4 + 12 = 16
|z| = r = 4
since cos cos θ and sin θ are positive ‘θ’ lies in 1st quadrant.
cos θ = \(\frac{1}{2}\), sin θ = \(\frac{√3}{2}\)
∴ θ = sin θ = \(\frac{π}{3}\) (or) θ = tan^-1 |\(\frac{y}{x}\)|
= tan^-1 |\(\frac{2√3}{2}\)|
= tan^-1 √3 = \(\frac{π}{3}\)
∴ argument = 2kπ + \(\frac{π}{3}\)
∴ Polar form is z = r (cos θ + i sin θ)
2 + 2i√3 = 4 (cos (2kπ + \(\frac{π}{3}\)) + i sin(2kπ +\(\frac{π}{3}\))) k ∈ z

(ii) 3 – i √3
Solution:
Let z = 3 – i √3 = r (cos θ + i sin θ)
equating real and imaginary parts
r cos θ = 3 (+ve)
r sin θ = -√3 (-ve)
r² cos² θ + r² sin² θ = (3)² + (-√3)²
r² = 9 + 3 = 12
|z| = r = 2√3
since cos cos θ positive and sin θ in -ve so lies in IV quadrant.
cos θ = \(\frac{√3}{2}\), sin θ = \(\frac{-1}{2}\), θ = \(\frac{-π}{6}\)
argument = 2kπ – \(\frac{π}{6}\), k ∈ Z
polar from z = r(cos θ + i sin θ)
3 – i√3 = 2√3 (cos (2kπ – \(\frac{π}{6}\)) + i sin(2kπ – \(\frac{π}{6}\))) k ∈ Z

(iii) -2 – i 2 = r (cos θ + i sin θ)
Solution:
Let z = -2 – i2 = r(cos θ + i sin θ)
equating real and imaginary parts
r cos θ = -2
r sin θ = -2
r² cos² θ + r² sin² θ = (-2)² + (-2)²
r² = 4 + 4 = 8
r² = 8
|z| = r = 2√2
cos θ = \(\frac{-2}{2√2}\) = \(\frac{-1}{√2}\), sin θ = \(\frac{-2}{2√2}\) = \(\frac{-1}{√2}\)
since cos θ and sin θ both are in -ve so lies in III quadrant.
argument = 2kπ – 3\(\frac{π}{4}\)
as θ = \(\frac{π}{4}\) – π = –\(\frac{3π}{4}\)
polar from z = r(cos θ + i sin θ)
-2 – i2 = 2√2 (cos (2kπ – \(\frac{3π}{4}\)) + i sin(2kπ – \(\frac{3}{4}\))) k ∈ Z

(iv) \(\frac{i-1}{cos{\frac{π}{3}}+isin{\frac{π}{3}}}\)
Solution:

Question 2.
Find the rectangular form of the complex numbers

Solution:

Solution:

Question 3.
\(\left(x_{1}+i y_{1}\right)\left(x_{2}+i y_{2}\right)\left(x_{3}+i y_{3}\right) \cdots\left(x_{n}+i y_{n}\right)=a+i b\), show that

Solution:
Let (x₁ + iy₁) (x₂ + iy₂) (x₃ + iy₃) …… (x_n + iy_n) = a + ib
Taking modulus
|(x₁ + iy₁) (x₂ + iy₂) (x₃ + iy₃) …… (x_n + iy_n)| = |a + ib|
|x₁ + iy₁| |x₂ + iy₂| |x₃ + iy₃| …… |x_n + iy_n| = |a + ib|
\(\sqrt{x_{1}^{2}+y_{1}^{2}} \sqrt{x_{2}^{2}+y_{2}^{2}} \sqrt{x_{3}^{2}+y_{3}^{2}} \ldots \sqrt{x_{n}^{2}+y_{n}^{2}}\) = \(\sqrt{a^2+b^2}\)
Squaring on both sides
\(\left(x_{1}^{2}+y_{1}^{2}\right)\left(x_{2}^{2}+y_{2}^{2}\right)\left(x_{3}^{2}+y_{3}^{2}\right) \ldots\left(x_{n}^{2}+y_{n}^{2}\right)\) = a² + b²
Hence proved

(ii) \(\sum_{r=1}^{n}\) tan^-1 (\(\frac{y_r}{x_r}\)) = tan^-1 (\(\frac{b}{a}\)) + 2kπ, k ∈ Z
Solution:
Let (x₁ + iy₁) (x₂ + iy₂) (x₃ + iy₃) …… (x_n + iy_n) = a + ib
Taking arguments
arg [(x₁ + iy₁) (x₂ + iy₂) (x₃ + iy₃) …… (x_n + iy_n)] = arg (a + ib)
arg (x₁ + iy₁) + arg(x₂ + iy₂) + arg (x₃ + iy₃) …… + arg(x_n + iy_n) = arg(a + ib)

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

Squaring on both sides
(1 + x)² + y² = (1 – x)² + y²
1 + 2x + x² + y² = 1 – 2x + x² +y²
x = 0
∴ z = 0 + iy = iy

∴ y = tan θ
hence z = iy
z = i tan θ

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) and
(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{π}{4}, then show that x² + y³ + 3x – 3y + 2 = 0.
Solution:

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

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 2 Complex Numbers Ex 2.6 Textbook Questions and Answers, Notes.

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

Question 1.
If 2 = 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:
Let z = x + iy

Squaring on both sides
x² + (y – 4)² =  x² + (y + 4)²
simplifying
We get y = 0
Which is 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

2y(1 – y) – x(2x + 1) = 0
⇒ 2y – 2y² – 2x² – x = 0
∴ The locus is 2x² + 2y² – 2y + x = 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
Solution:
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
Solution:

x² + y² = 1, x² + y² = -1 which cannot be true.
∴ x² + y² = 1

Question 4.
Show that the following equations represent a circle, and, find its centre and radius.
(i) |z – 2 – i| = 3
Solution:
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
Solution:
Let z = x + iy
|x + iy – 4| – 16
|(x – 4) + iy| = 16
\(\sqrt{(x – 4)² + y²}\) = 16
∴ Squaring on both sides
(x – 4)² + y² = 256
x² – 8x + 16 + y² – 256 = 0
x² + y² – 8x – 240 = 0
The locus of the point is a circle.

(ii) |z – 4|² – |z – 1|² = 16.
Solution:
|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
The locus of the point is a straight line.

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 2 Complex Numbers Ex 2.5 Textbook Questions and Answers, Notes.

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

Question 1.
Find the modulus of the following complex numbers.
(i) \(\frac{2i}{3+4i}\)
Solution:

(ii) \(\frac{2-i}{1+i}+\frac{1-2 i}{1-i}\)
Solution:

Modulus of z = |z| = \(\sqrt{4+4}\)
= √8
= 2√2

(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 real number.
Solution:
Given |z₁| = |z₂| = 1 and z₁ z₂ ≠ 1
|z₁|² = 1 |z₂|² = 1
z₁ \(\bar{z}_{1}\) = 1 similarly z₂ \(\bar{z}_{2}\) = 1

Since z = \(\bar{z}\), it is a real number.

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:
given |z| = 3
|z + 6 – 8i| ≤ |z| + |6 – 8i|
= 3 + \(\sqrt{6^2+8^2}\)
= 3 + \(\sqrt{100}\)
= 3 + 10 = 13
∴ |z + 6 – 8i| ≤ 13 ……….. (1)
|z + 6 – 8i| ≥ ||z| – |-6 + 8i||
= |3 – 10|
= |-7| = 7
∴ |z + 6 – 8i| ≥ 7 ………… (2)
from 1 and 2
we get 7 ≤ |z + 6 – 8i| ≤ 13
hence proved.

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 |z| = 2 show that the 8 ≤ |z + 6 + 8i| ≤ 12
Solution:
Given |z| = 2
|z + 6 + 8i| = |z| + |6 + 8i|
= 2 + \(\sqrt{6^2+8^2}\)
= 2 + \(\sqrt{100}\)
= 2 + 10
= 12
∴ |z + 6 + 8i| ≤ 12 ……….. (1)
|z + 6 + 8i| ≥ ||z| – |-6 – 8i||
= |2 – 10|
= |-8|
= 8
|z + 6 + 8i| ≥ 8 ………… (2)
From 1 and 2 we get
8 ≤ |z + 6 + 8i| ≤ 12
Hence proved

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

Hence proved.

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
Aliter:
Given the area of triangle = 50 sq. unit

x² + y² = 100
|z|² = 100
|z| = 10

Question 9.
Show that the equation z³ + 2 \(\bar {z}\) = 0 has five solutions.
Solution:
Given z³ + 2 \(\bar {z}\) = 0
z³ = -2 \(\bar {z}\)
|z³| = |-2| |\(\bar {z}\)|
|z|³ = 2|z| [∵ |z| = |\(\bar {z}\)|
|z|³ – 2 |z| = 0
|z| [|z|² – 2] = 0
|z| = 0 |z|² = 2
z\(\bar {z}\) = 2
z = \(\frac{2}{\bar {z}}\) = ± √2 [∵ \(\bar {z}\) = \(\frac{-z^3}{2}\) ]
z = \(\frac{2}{(\frac{z^3}{-2})}\)
z⁴ = 4
It has 4 non zero solutions.
∴ Including z = 0 we have 5 solutions.

Question 10.
Find the square roots of
(i) 4 + 3i
Solution:
|4 + 3i| = \(\sqrt {4^2+3^2}\) = \(\sqrt {16+9}\)
\(\sqrt {25}\) = 5
Let \(\sqrt {4+3i}\) = a + ib
squaring on both sides
4 + 3i = (a + ib)²
4 + 3i = (a² – b²) + 2 jab
Equating real and imaginary parts
a² – b² = 4, 2ab = 3
(a² + b²)² = (a² – b²)² + 4a² b²
= (4)² + (3)²
= 16 + 9 = 25
∴ a² + b² = 5
Solving a² – b² = 4 and a² + b² = 5.
we get a² = \(\frac {9}{2}\) , b² = \(\frac {1}{2}\)
a = ±\(\frac {3}{√2}\) and b = ±\(\frac {1}{√2}\)
∴ \(\sqrt {4 + 3i}\) = a + ib
= ±(\(\frac {3}{√2}\) + ±\(\frac {i}{√2}\))
Aliter:
Square root of 4 + 3i
formula method

(ii) -6 + 8i
Solution:
Let \(\sqrt {-6 + 8i}\) = a + ib
Squaring on both sides
-6 + 8i = (a + ib)²
-6 + 8i = a² – b² + 2iab
Equating real and imaginary parts
a² – b² = -6 and 2ab = 8
Now (a² + b²)² = (a² – b²)² + 4a²b²
= (-6)² + (8)²
= 36 + 64 = 100
∴ a + b² = 10
Solving a² – b² = -6 and a² + b² = 10
we get 2a² = 4, b² = 8
a² = 2, b² = ±2√2
a = ±√2
∴ \(\sqrt {-6 + 8i}\) = ±√2 ± i 2√2
= ±(√2 + i 2√2)
Aliter:
square root of -6 + 8i
let a + ib = -6 + 8i
a = -6, b = 8

(iii) -5 – 12i
Solution:
Let \(\sqrt{-5-12 i}\) = a + ib
Squaring on both sides
-5 – 12i = (a+ib)²
-5 – 12i = a² – b² + 2iab
Equating real and imaginary parts
a² – b² = -5, 2ab = -12
(a² + b²)² = (a² – b²)² + 4a²b²
= (-5)² + (-12)² = 169
∴ a² + b² = 13
Solving a²- b² = -5 and a² + b² = 13
we get a² = 4, b² = 9
a = ±2, b = ±3
Since 2ab = -12 < 0, a, b are of opposite signs.
∴ When a = ±2, b = ±3
Now \(\sqrt{-5-12 i}\) = ± (2 – 3i)
Aliter
Square root of -5 – 12i
Let a + ib = -5 – 12i
a = -5, b = -12

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 2 Complex Numbers Ex 2.4 Textbook Questions and Answers, Notes.

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

Question 1.
Write the following in the rectangular form:
(i) \(\overline { (5+9i)+(2-4i) } \)
Solution:
\(\overline { (5+9i)+(2-4i) } \)
= \(\overline {(5+9i)} \) + \(\overline {(2-4i)} \)
= 5 – 9i + 2 + 4i
= 7 – 5i

(ii) \(\frac {10-5i}{6+2i} \)
Solution:

(iii) \(\overline {3i} + \frac{2}{2-i}\)
Solution:

Question 2.
If z = x + iy, find the following in rectangular form.
(i) Re(\(\frac {1}{z} \))
Answer:

(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
z₁ z₂ = (2 – i) (-4 + 3i)
= -8 + 3 + 4i + 6i
= -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

Question 5.
Prove the following properties:
(i) z is real if and only if z = \(\overline {z}\)
Solution:
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) Re(z) = \(\frac{z+\bar{z}}{2}\) and Im(z) = \(\frac{z-\bar{z}}{2i}\)
Solution:
let z = x + iy
\(\overline {z}\) = x – iy

Hence proved

Question 6.
Find the least value of the positive integer n for which (√3 + i)ⁿ (i) real, (ii) purely imaginary.
Solution:
Given (√3 + i)ⁿ
= (√3)² + 2i √3 + (i)²
= 3 + 2i √3 – 1
= 2 + 2i √3
= 2(1 + i√3)
put n = 3 or 4 or 5
then real part is not possible

which is purely real ∴ n = 6

(ii) (√3 + i)ⁿ

which is purely imaginary
∴ n = 3

Question 7.
Show that
(i) (2 + i√3)¹⁰ – (2 – i√3)¹⁰ is purely imaginary.
Solution:
Let z = (2 + i√3)¹⁰ – (2 – i√3)¹⁰
Let Z

= (2 – i√3)¹⁰ – (2 + i√3)¹⁰
= -[(z + i√3)¹⁰ – (2 – i√3)¹⁰]
= -z
(2 + i√3)¹⁰ – (2 – i√3)¹⁰ is purely imaginary

(ii) \(\left(\frac{19-7 i}{9+i}\right)^{12}\) + \(\left(\frac{20-5 i}{7-6 i}\right)^{12}\) is real
Solution:

∴ z is real.