# Samacheer Kalvi 12th Maths Guide Chapter 2 Complex Numbers Ex 2.5

## 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 = 25 = 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 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| = 1 ⇒ z1 $$\bar{z}$$1 = 1
z1 = $$\frac{1}{\bar{z}_{1}}$$
similarly z2 = $$\frac{1}{\bar{z}_{2}}$$

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 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|² = 1, |z2|² = 4, |z3|² = 9
z1 $$\bar {z_1}$$ = 1, z2 $$\bar {z_2}$$ = 4, z3 $$\bar {z_3}$$ = 9
$$\bar {z_1}$$ = $$\frac{1}{z_1}$$ $$\bar {z_2}$$ = $$\frac{4}{z_2}$$, $$\bar {z_3}$$ = $$\frac{9}{z_3}$$

⇒ |9z1 z2 + 4 z1 z3 + z2 z3| = 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
z4 ± 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)

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