# Samacheer Kalvi 12th Maths Solutions 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) $$\frac{2 i}{3+4 i}$$
(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
Solution:
|z| = |$$\sqrt{1^2+1^2}$$|10 [∵ |zn| = |z|n]
= (√2)10 = 25 = 32

(iv) 2i(3 – 4i)(4 – 3i)
Solution:
= |2i||3 – 4i||4 – 3i|
= 2i(12 – 9i – 16i – 12)
= 2i(-25i) = 50
∴ |z| = 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:
Let z1 = 1 + i
z2 = 10 – 8i
z3 = 11 + 6i
|z1– z2| = |1 + i – 10 + 8i|
= |-9 + 9i|
= $$\sqrt{(-9^2)+(9^2)}$$
= $$\sqrt{162}$$ = 9√2
|z1– z3| = |1 + i – 11 – 6i|
= |-10 – 5i|= $$\sqrt{(-10^2) + (-5^2)}$$
= $$\sqrt{100+25}$$
= $$\sqrt{125}$$
= $$\sqrt{5×25}$$ = 5√5
∴ 5√5 < 9√2
∴ |z1– z2| > |z1– z3|
∴ 11 + 6i is closest to 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:
given |z| = 1
Now |z² – 3| ≤ |z²| + |-3|
= |z|² + 3
= 1 + 3 = 4
∴ |z² – 3| ≤ 4 ……….. (1)
|z² – 3| ≥ ||z²|-|-3||
= |1 – 3| = |-2| = 2
|z² – 3| ≥ 2 ………. (2)
From 1 and 2
2 ≤ |z² – 3| ≤ 4
Hence Proved.

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 vertices are z = x + iy.
iz = i(x + iy) = -y + ix
z + iz = (x – y) + i(x + y)
the points representing the complex numbers z, iz and z + iz are (x, y), (-y, x) and ((x – y), (x + y)) respectively.
given area of triangle = 50 square units
|$$\frac {1}{2}$$ [x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)]| = 50
|x(x – x – y) -y(x + y – y) + (x – y) (y – x)| = 100
|-xy – xy + 2xy – x² – y²| = 100
x² + y² = 100
|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: