## 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:

= 2√2

Samacheer Kalvi 12th Maths Guide

(iii) (1 – i)^{10}

Solution:

|(1 – i)^{10}| = (|1 – i|)^{10} = (\(\sqrt{1+1}\))^{10}

= (√2)^{10} = 2^{5} = 32

(iv) 2i(3 – 4i)(4 – 3i)

Solution:

= |2i||3 – 4i||4 – 3i|

= (\(\sqrt{0+4}\)) (\(\sqrt{9+16}\)) (\(\sqrt{16+9}\))

= 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}| = 1 ⇒ z_{1} \(\bar{z}\)_{1} = 1

z_{1} = \(\frac{1}{\bar{z}_{1}}\)

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

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:

Let A be (10 – 8i) ⇒ (10, -8)

B be (11 + 6i) ⇒ (11, 6)

C be (1 + i) = (1, 1)

CA = |(1 + i) – (10 – 8i)|

= |- 9 + 9i| = \(\sqrt{81+81}\)

= \(\sqrt{162}\) = 9√2

CB = |(1 + i) – (11 + 6i)

= |-10 – 5i|= \(\sqrt{100 + 25}\)

= \(\sqrt{125}\)

∴ The point B is closer to C, than A.

Question 4.

If |z| = 3, show that 7 ≤ |z + 6 – 8i| ≤ 13.

Solution:

To prove 7 ≤ |z + 6 – 8i| ≤ 13

|z + 6 – 8i| ≥ |z| + |16 + 8i| = 3 + \(\sqrt{36 + 64}\) = 13

Also |z + 6 – 8i| ≤ ||z| – (6 + 8i)| = |3 – \(\sqrt{36 + 64}\)|

= |3 – 10| = 7

∴ 7 ≤ |z + 6 – 8i| ≤ 13

Question 5.

If |z| = 1, show that 2 ≤ |z² – 3| ≤ 4.

Solution:

|z² – 3| ≤ |z²| + |-3| = 1 + 3 = 4

Also |z² – 3| ≥ ||z²|-|-3|| = |1-3| = 2

Hence 2 ≤ |z² – 3| ≤ 4

Question 6.

If |z – \(\frac{2}{z}\)|= 2 show that the greatest and least value of |z| are √3 + 1 and √3 – 1 respectively.

Solution:

Given |z – \(\frac{2}{z}\)|= z

|z² – 2|= 2|z|

|z²| ≤ 2|z| + |-2|

≤ 2 |z| + 2

|z²|- 2|z| + 1 ≤ 2 + 1

(|z – 1|)² ≤ 3

|z| – 1 ≤ √3

|z| ≤ √3 + 1

Similarly |z| ≥ √3 – 1

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}|² = 1, |z_{2}|² = 4, |z_{3}|² = 9

z_{1} \(\bar {z_1}\) = 1, z_{2} \(\bar {z_2}\) = 4, z_{3} \(\bar {z_3}\) = 9

\(\bar {z_1}\) = \(\frac{1}{z_1}\) \(\bar {z_2}\) = \(\frac{4}{z_2}\), \(\bar {z_3}\) = \(\frac{9}{z_3}\)

⇒ |9z_{1} z_{2} + 4 z_{1} z_{3} + z_{2} z_{3}| = 6

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 vertices are z, iz and z + iz.

Since it side is z and other side is iz, it shows that they are perpendicular and z + iz is the hypotenuse.

Area of ΔAOB = \(\frac{1}{2}\) |z| |zi|

⇒ \(\frac{1}{2}\) |z| |z| = 50 (given)

⇒ |z|² = 100

|z| = 10 units

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| |\(\bar {z}\)|

= 2|z| since |\(\bar {z}\)| = |z|

|z|³ – 2 |z| = 0

|z|(|z|² – 2) = 0

⇒ |z| = 0 |z|² = 2

⇒ z = 0 is a solution |z| ± √2

⇒ z \(\bar {z}\) = ± √2

\(\bar {z}\) = ± \(\frac {√2}{z}\)

But z³ + 2 \(\bar {z}\) = 0

∴ z³ + 2(±\(\frac {√2}{z}\)) = 0

z^{4} ± 2 √2 z = 0

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| = \(\sqrt {16±9}\) = 5

(ii) -6 + 8i

Solution:

To find

(Here also b = 8 is + ve)

±\((\sqrt{\frac{4}{2}}+i \sqrt{\frac{16}{2}})\) = ± (2 ± 2√2 i)

(iii) -5 – 12i

Solution:

To find \(\sqrt{-5-12 i}\) = |-5 – 12i|

= \(\sqrt{25+144}\) = 13

Using the formula

= ±(2 – i3)