Samacheer Kalvi Class 12 Maths Solutions Chapter 2 Complex Numbers Exercise 2.3

Question 1. If \( z_1 = 1 - 3i \), \( z_2 = -4i \), and \( z_3 = 5 \), show that
(i) \( (z_1 + z_2) + z_3 = z_1 + (z_2 + z_3) \)
(ii) \( (z_1 z_2)z_3 = z_1(z_2 z_3) \)
Answer:
Given complex numbers are: \( z_1 = 1 - 3i \), \( z_2 = -4i \), and \( z_3 = 5 \).

(i) To show: \( (z_1 + z_2) + z_3 = z_1 + (z_2 + z_3) \)
First, calculate the left-hand side (LHS):
\( (z_1 + z_2) = (1 - 3i) + (-4i) = 1 - 3i - 4i = 1 - 7i \)
Then, \( (z_1 + z_2) + z_3 = (1 - 7i) + 5 = 1 + 5 - 7i = 6 - 7i \). (1)

Next, calculate the right-hand side (RHS):
\( (z_2 + z_3) = (-4i) + 5 = 5 - 4i \)
Then, \( z_1 + (z_2 + z_3) = (1 - 3i) + (5 - 4i) = 1 + 5 - 3i - 4i = 6 - 7i \). (2)
From (1) and (2), we see that the LHS is equal to the RHS.
Thus, \( (z_1 + z_2) + z_3 = z_1 + (z_2 + z_3) \) is proved. This shows that addition of complex numbers is associative.

(ii) To show: \( (z_1 z_2)z_3 = z_1(z_2 z_3) \)
First, calculate the left-hand side (LHS):
\( z_1 z_2 = (1 - 3i)(-4i) = 1(-4i) - 3i(-4i) = -4i + 12i^2 = -4i + 12(-1) = -12 - 4i \)
Then, \( (z_1 z_2)z_3 = (-12 - 4i)(5) = -12(5) - 4i(5) = -60 - 20i \). (1)

Next, calculate the right-hand side (RHS):
\( z_2 z_3 = (-4i)(5) = -20i \)
Then, \( z_1(z_2 z_3) = (1 - 3i)(-20i) = 1(-20i) - 3i(-20i) = -20i + 60i^2 = -20i + 60(-1) = -60 - 20i \). (2)
From (1) and (2), we see that the LHS is equal to the RHS.
Thus, \( (z_1 z_2)z_3 = z_1(z_2 z_3) \) is proved. This means multiplication of complex numbers is also associative.
In simple words: We checked if adding or multiplying complex numbers in different orders gives the same answer. We did this for both addition and multiplication by grouping numbers differently, and in both cases, the final result was the same, proving the associative property.

🎯 Exam Tip: Remember to handle the \( i^2 = -1 \) term carefully in multiplication, as it often converts imaginary parts into real numbers, changing the complex number's structure.

Question 2. If \( z_1 = 3 \), \( z_2 = 7i \), and \( z_3 = 5 + 4i \), show that
(i) \( z_1(z_2 + z_3) = z_1 z_2 + z_1 z_3 \)
(ii) \( (z_1 + z_2)z_3 = z_1 z_3 + z_2 z_3 \)
Answer:
Given complex numbers are: \( z_1 = 3 \), \( z_2 = 7i \), and \( z_3 = 5 + 4i \).

(i) To show: \( z_1(z_2 + z_3) = z_1 z_2 + z_1 z_3 \)
First, calculate the left-hand side (LHS):
\( z_2 + z_3 = 7i + (5 + 4i) = 5 + 7i + 4i = 5 + 11i \)
Then, \( z_1(z_2 + z_3) = 3(5 + 11i) = 3(5) + 3(11i) = 15 + 33i \). (1)

Next, calculate the right-hand side (RHS):
\( z_1 z_2 = 3(7i) = 21i \)
\( z_1 z_3 = 3(5 + 4i) = 3(5) + 3(4i) = 15 + 12i \)
Then, \( z_1 z_2 + z_1 z_3 = 21i + (15 + 12i) = 15 + 21i + 12i = 15 + 33i \). (2)
From (1) and (2), we see that the LHS is equal to the RHS.
Thus, \( z_1(z_2 + z_3) = z_1 z_2 + z_1 z_3 \) is proved. This demonstrates the distributive property of multiplication over addition for complex numbers.

(ii) To show: \( (z_1 + z_2)z_3 = z_1 z_3 + z_2 z_3 \)
First, calculate the left-hand side (LHS):
\( z_1 + z_2 = 3 + 7i \)
Then, \( (z_1 + z_2)z_3 = (3 + 7i)(5 + 4i) \)
\( = 3(5) + 3(4i) + 7i(5) + 7i(4i) \)
\( = 15 + 12i + 35i + 28i^2 \)
\( = 15 + 47i + 28(-1) \)
\( = 15 + 47i - 28 \)
\( = -13 + 47i \). (1)

Next, calculate the right-hand side (RHS):
\( z_1 z_3 = 3(5 + 4i) = 15 + 12i \)
\( z_2 z_3 = 7i(5 + 4i) = 7i(5) + 7i(4i) = 35i + 28i^2 = 35i - 28 = -28 + 35i \)
Then, \( z_1 z_3 + z_2 z_3 = (15 + 12i) + (-28 + 35i) \)
\( = 15 - 28 + 12i + 35i \)
\( = -13 + 47i \). (2)
From (1) and (2), we see that the LHS is equal to the RHS.
Thus, \( (z_1 + z_2)z_3 = z_1 z_3 + z_2 z_3 \) is proved. This is another way to express the distributive property for complex numbers.
In simple words: We checked the distributive property, which means you can multiply a number by a sum, or multiply it by each part of the sum and then add, and get the same result. We proved this for complex numbers in two different ways.

🎯 Exam Tip: When proving properties, always clearly label LHS and RHS calculations. Carefully combine real and imaginary parts in the final step to avoid errors.

Question 3. If \( z_1 = 2 + 5i \), \( z_2 = -3 - 4i \), and \( z_3 = 1 + i \), find the additive and multiplicative inverse of \( z_1, z_2 \), and \( z_3 \).
Answer:
(i) For \( z_1 = 2 + 5i \):
The additive inverse of \( z_1 \) is \( -z_1 \). This means we change the sign of both the real and imaginary parts.
\( -z_1 = -(2 + 5i) = -2 - 5i \)
The multiplicative inverse of \( z_1 \) is \( z_1^{-1} \). We know that \( z_1 z_1^{-1} = 1 \).
Let \( z_1^{-1} = u + iv \).
\( (2 + 5i)(u + iv) = 1 \)
\( 2u + 2iv + 5iu + 5i^2v = 1 \)
\( 2u - 5v + i(5u + 2v) = 1 + 0i \)
Comparing the real and imaginary parts, we get a system of two linear equations:
1) \( 2u - 5v = 1 \)
2) \( 5u + 2v = 0 \)
To solve for \( u \) and \( v \), we can multiply equation (1) by 2 and equation (2) by 5:
\( 4u - 10v = 2 \)
\( 25u + 10v = 0 \)
Adding these two new equations:
\( (4u - 10v) + (25u + 10v) = 2 + 0 \)
\( 29u = 2 \)
\( u = \frac{2}{29} \)
Substitute \( u \) back into equation (2):
\( 5(\frac{2}{29}) + 2v = 0 \)
\( \frac{10}{29} + 2v = 0 \)
\( 2v = -\frac{10}{29} \)
\( v = -\frac{5}{29} \)
So, the multiplicative inverse \( z_1^{-1} = \frac{2}{29} - i\frac{5}{29} = \frac{1}{29}(2 - 5i) \).

(ii) For \( z_2 = -3 - 4i \):
The additive inverse of \( z_2 \) is \( -z_2 \).
\( -z_2 = -(-3 - 4i) = 3 + 4i \)
The multiplicative inverse of \( z_2 \) is \( z_2^{-1} \). Let \( z_2^{-1} = u + iv \).
\( (-3 - 4i)(u + iv) = 1 \)
\( -3u - 3iv - 4iu - 4i^2v = 1 \)
\( -3u + 4v + i(-4u - 3v) = 1 + 0i \)
Comparing the real and imaginary parts:
1) \( -3u + 4v = 1 \)
2) \( -4u - 3v = 0 \)
To solve for \( u \) and \( v \), multiply equation (1) by 3 and equation (2) by 4:
\( -9u + 12v = 3 \)
\( -16u - 12v = 0 \)
Adding these two new equations:
\( (-9u + 12v) + (-16u - 12v) = 3 + 0 \)
\( -25u = 3 \)
\( u = -\frac{3}{25} \)
Substitute \( u \) back into equation (2):
\( -4(-\frac{3}{25}) - 3v = 0 \)
\( \frac{12}{25} - 3v = 0 \)
\( 3v = \frac{12}{25} \)
\( v = \frac{4}{25} \)
So, the multiplicative inverse \( z_2^{-1} = -\frac{3}{25} + i\frac{4}{25} = \frac{1}{25}(-3 + 4i) \).

(iii) For \( z_3 = 1 + i \):
The additive inverse of \( z_3 \) is \( -z_3 \).
\( -z_3 = -(1 + i) = -1 - i \)
The multiplicative inverse of \( z_3 \) is \( z_3^{-1} \). We can find this by multiplying by the conjugate.
\( z_3^{-1} = \frac{1}{1 + i} = \frac{1}{1 + i} \times \frac{1 - i}{1 - i} = \frac{1 - i}{1^2 - i^2} = \frac{1 - i}{1 - (-1)} = \frac{1 - i}{2} = \frac{1}{2} - i\frac{1}{2} \).
In simple words: The additive inverse of a complex number is found by just changing the signs of both its real and imaginary parts. The multiplicative inverse is a bit trickier; it's the number that, when multiplied by the original number, gives 1. We find it either by solving a system of equations or by using the conjugate.

🎯 Exam Tip: Always remember that \( i^2 = -1 \) and that \( \frac{1}{a+bi} = \frac{a-bi}{a^2+b^2} \) is a useful shortcut for finding the multiplicative inverse quickly.