Samacheer Kalvi Class 12 Maths Solutions Chapter 2 Complex Numbers Exercise 2.2

Question 1. Evaluate the following if \( z = 5 - 2i \) and \( w = -1 + 3i \)
(i) \( z + w \)
(ii) \( z - iw \)
(iii) \( 2z + 3w \)
(iv) \( zw \)
(v) \( z^2 + 2zw + w^2 \)
(vi) \( (z + w)^2 \)
Answer:
(i) First, we are given the complex numbers \( z = 5 - 2i \) and \( w = -1 + 3i \).
To find \( z + w \), we add the real parts together and the imaginary parts together:
\( z + w = (5 - 2i) + (-1 + 3i) \)
\( = (5 - 1) + (-2i + 3i) \)
\( = 4 + i \)
(ii) To find \( z - iw \), we first calculate \( iw \):
\( iw = i(-1 + 3i) = -i + 3i^2 = -i + 3(-1) = -3 - i \)
Now, we subtract \( iw \) from \( z \):
\( z - iw = (5 - 2i) - (-3 - i) \)
\( = 5 - 2i + 3 + i \)
\( = (5 + 3) + (-2i + i) \)
\( = 8 - i \)
(iii) To find \( 2z + 3w \), we first multiply \( z \) by 2 and \( w \) by 3:
\( 2z = 2(5 - 2i) = 10 - 4i \)
\( 3w = 3(-1 + 3i) = -3 + 9i \)
Then, we add the results:
\( 2z + 3w = (10 - 4i) + (-3 + 9i) \)
\( = (10 - 3) + (-4i + 9i) \)
\( = 7 + 5i \)
(iv) To find \( zw \), we multiply the two complex numbers:
\( zw = (5 - 2i)(-1 + 3i) \)
\( = 5(-1) + 5(3i) - 2i(-1) - 2i(3i) \)
\( = -5 + 15i + 2i - 6i^2 \)
We know that \( i^2 = -1 \), so we substitute this value:
\( = -5 + 17i - 6(-1) \)
\( = -5 + 17i + 6 \)
\( = 1 + 17i \)
(v) To find \( z^2 + 2zw + w^2 \), we can recognize this as the expansion of \( (z + w)^2 \).
From part (i), we already found that \( z + w = 4 + i \).
So, we calculate the square of this result:
\( (z + w)^2 = (4 + i)^2 \)
\( = 4^2 + 2(4)(i) + i^2 \)
\( = 16 + 8i - 1 \)
\( = 15 + 8i \)
(vi) From part (v), we already calculated \( (z + w)^2 \). This expression represents the square of the sum of the complex numbers.
\( (z + w)^2 = 15 + 8i \)
In simple words: When adding or subtracting complex numbers, combine their real parts and imaginary parts separately. For multiplication, use the distributive property, remembering that \( i^2 = -1 \). If an expression looks like \( (a+b)^2 \), you can often use a previously calculated sum.

🎯 Exam Tip: Always group the real terms and imaginary terms carefully during addition or subtraction. For multiplication, remember to replace \( i^2 \) with \( -1 \) to simplify the expression completely.

Question 2. Given the complex number \( z = 2 + 3i \), represent the complex numbers in the Argand diagram.
(i) \( z = 2 + 3i \)
(ii) \( z, -iz \) and \( z - iz \)
Answer:
(i) First, we calculate the values for \( iz \) and \( z + iz \):
\( z = 2 + 3i \)
To find \( iz \), we multiply \( z \) by \( i \):
\( iz = i(2 + 3i) \)
\( = 2i + 3i^2 \)
\( = 2i + 3(-1) \)
\( = -3 + 2i \)
To find \( z + iz \), we add \( z \) and \( iz \):
\( z + iz = (2 + 3i) + (-3 + 2i) \)
\( = (2 - 3) + (3i + 2i) \)
\( = -1 + 5i \)
Now, we plot these points on the Argand diagram:
\( z \) corresponds to the point \( (2, 3) \)
\( iz \) corresponds to the point \( (-3, 2) \)
\( z + iz \) corresponds to the point \( (-1, 5) \)

(ii) First, we calculate the values for \( -iz \) and \( z - iz \):
\( z = 2 + 3i \)
To find \( -iz \), we multiply \( z \) by \( -i \):
\( -iz = -i(2 + 3i) \)
\( = -2i - 3i^2 \)
\( = -2i - 3(-1) \)
\( = 3 - 2i \)
To find \( z - iz \), we subtract \( -iz \) from \( z \):
\( z - iz = (2 + 3i) + (3 - 2i) \)
\( = (2 + 3) + (3i - 2i) \)
\( = 5 + i \)
Now, we plot these points on the Argand diagram:
\( z \) corresponds to the point \( (2, 3) \)
\( -iz \) corresponds to the point \( (3, -2) \)
\( z - iz \) corresponds to the point \( (5, 1) \)

In simple words: An Argand diagram is like a normal graph but for complex numbers. The horizontal line is for the real part and the vertical line is for the imaginary part. We find the coordinates for each complex number and then mark them on this special graph.

🎯 Exam Tip: When plotting complex numbers, remember that \( (x, y) \) corresponds to \( x + iy \). Multiplying by \( i \) rotates a complex number \( 90^\circ \) counter-clockwise on the Argand plane.

Question 3. Find the values of the real numbers x and y, if the complex numbers \( (3 - i)x - (2 - i) y + 2i + 5 \) and \( 2x + (-1 + 2i)y + 3 + 2i \) are equal.
Answer: We are given that two complex numbers are equal. We will first expand and group the real and imaginary parts of each expression.
Given: \( (3 - i)x - (2 - i) y + 2i + 5 = 2x + (-1 + 2i)y + 3 + 2i \)
Expand the left side:
\( 3x - ix - 2y + iy + 2i + 5 \)
Group real and imaginary parts on the left side:
\( (3x - 2y + 5) + i(-x + y + 2) \)
Expand the right side:
\( 2x - y + 2yi + 3 + 2i \)
Group real and imaginary parts on the right side:
\( (2x - y + 3) + i(2y + 2) \)
Since the two complex numbers are equal, their real parts must be equal, and their imaginary parts must be equal.
Equating the real parts:
\( 3x - 2y + 5 = 2x - y + 3 \)
Rearrange the terms to form an equation:
\( 3x - 2x - 2y + y = 3 - 5 \)
\( x - y = -2 \) ...... (1)
Equating the imaginary parts:
\( -x + y + 2 = 2y + 2 \)
Rearrange the terms to form another equation:
\( -x + y - 2y = 2 - 2 \)
\( -x - y = 0 \)
\( x + y = 0 \) ...... (2)
Now we have a system of two linear equations with two variables:
(1) \( x - y = -2 \)
(2) \( x + y = 0 \)
Add equation (1) and equation (2):
\( (x - y) + (x + y) = -2 + 0 \)
\( x - y + x + y = -2 \)
\( 2x = -2 \)

\( \implies x = -1 \)
Substitute the value of \( x = -1 \) into equation (2):
\( -1 + y = 0 \)

\( \implies y = 1 \)
Thus, the values of the real numbers are \( x = -1 \) and \( y = 1 \).
In simple words: To find unknown numbers in equal complex expressions, separate each side into its real and imaginary parts. Then, set the real parts equal to each other and the imaginary parts equal to each other. This creates two simpler equations that you can solve to find the unknowns.

🎯 Exam Tip: When equating complex numbers, ensure all real terms are grouped together and all imaginary terms are grouped together on both sides before setting them equal. Any \( i^2 \) term must be converted to \( -1 \).