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 z_{1} and z_{2}, such that |z_{1}| = |z_{2}| = 1 and z_{1} z_{2} ≠ -1, then show that \(\frac{z_{1}+z_{2}}{1+z_{1} z_{2}}\) is real number.

Solution:

Given |z_{1}| = |z_{2}| = 1 and z_{1} z_{2 }≠ 1

|z_{1}|² = 1 |z_{2}|² = 1

z_{1} \(\bar{z}_{1}\) = 1_{ }similarly z_{2} \(\bar{z}_{2}\) = 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

||z_{1}| – |z_{2}|| ≤ |z_{1} + z_{2}| ≤ |z_{1}| + |z_{2}|

||z|^{2} – |-3|| ≤ |z^{2} – 3| ≤ |z|^{2} + |-3|

|1 – 3| ≤ |z^{2} – 3| ≤ 1 + 3

2 ≤ |z^{2} – 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 z_{1} z_{2} and z_{3} are three complex numbers such that |z_{1}| = 1, |z_{1}| = 2, |z_{3}| = 3 and |z_{1} + z_{2} + z_{3}| = 1 show that |9z_{1} z_{2} + 4z_{1} z_{3} + z_{2} z_{3}| = 6.

Solution:

|z_{1}| = 1, |z_{1}| = 2, |z_{3}| = 3

|z_{1} + z_{2} + z_{3}| = 1

Now |9z_{1} z_{2} + 4z_{1} z_{3} + z_{2} z_{3}|

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 = \(\frac{2}{(\frac{z^3}{-2})}\)

z^{4} = 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