1.4 Q-4 | Complex Numbers | Mathematics 11

Exercise 1.4 — Question 4

This exercise covers the equality of complex numbers, basic algebraic operations, and solving simultaneous linear equations with complex coefficients.

Key Concepts

1. Equality of Complex Numbers (M-11-A-03)

Two complex numbers $z_{1} = a + ib$ and $z_{2} = c + i d$ are equal if and only if: $a = c and b = d$ That is, their real parts are equal and their imaginary parts are equal.

Example: If $(2 x - 1) + (y + 3) i = 5 + 2 i$ , then:

Real parts: $2 x - 1 = 5 \Rightarrow x = 3$
Imaginary parts: $y + 3 = 2 \Rightarrow y = - 1$

2. Basic Operations on Complex Numbers (M-11-A-04)

For $z_{1} = a + ib$ and $z_{2} = c + i d$ :

Operation	Result
Addition	$(a + c) + i (b + d)$
Subtraction	$(a - c) + i (b - d)$
Multiplication	$(a c - b d) + i (a d + b c)$
Division	$\frac{( a c + b d ) + i ( b c - a d )}{c ^{2} + d ^{2}}$

Modulus of a product: $∣ z_{1} z_{2} \dots z_{n} ∣ = ∣ z_{1} ∣∣ z_{2} ∣ \dots ∣ z_{n} ∣$

Squaring both sides, if $(x_{1} + i y_{1}) (x_{2} + i y_{2}) \dots (x_{n} + i y_{n}) = a + ib$ , then: $(x_{1}^{2} + y_{1}^{2}) (x_{2}^{2} + y_{2}^{2}) \dots (x_{n}^{2} + y_{n}^{2}) = a^{2} + b^{2}$

Example: $(1 + 2 i) (3 - i) = 3 - i + 6 i - 2 i^{2} = 3 + 5 i + 2 = 5 + 5 i$ .

Check using modulus: $(1^{2} + 2^{2}) (3^{2} + 1^{2}) = 5 \times 10 = 50 = 5^{2} + 5^{2}$ ✓

3. Simultaneous Linear Equations with Complex Coefficients (M-11-A-07)

To solve a system of simultaneous equations with complex coefficients, use elimination or substitution.

Example: Solve $z + i w = 1 + 2 i \dots (1)$ and $i z - w = 3 \dots (2)$

Method — Elimination:

Multiply equation (1) by $i$ : $i z + i^{2} w = i (1 + 2 i) \Rightarrow i z - w = i + 2 i^{2} = - 2 + i \dots (3)$

Subtract equation (2) from (3): $(i z - w) - (i z - w) = (- 2 + i) - 3$ $0 = - 5 + i (contradiction — system is inconsistent for these values)$

Note: Not every system has a solution. If elimination leads to a contradiction (e.g. $0 = k$ where $k \neq = 0$ ), the system is inconsistent (no solution).

Example of a consistent system: Solve $z + i w = 2 + i \dots (1)$ and $i z + w = 1 + 3 i \dots (2)$

From (1): $z = 2 + i - i w$ . Substitute into (2): $i (2 + i - i w) + w = 1 + 3 i$ $2 i + i^{2} - i^{2} w + w = 1 + 3 i$ $2 i - 1 + w + w = 1 + 3 i$ $- 1 + 2 i + 2 w = 1 + 3 i$ $2 w = 2 + i \Rightarrow w = 1 + \frac{i}{2}$

Substitute back: $z = 2 + i - i (1 + \frac{i}{2}) = 2 + i - i + \frac{1}{2} = \frac{5}{2}$

General Method (substitution of $z = x + i y$ ):

If the unknowns are complex, write $z = x + i y$ and $w = u + i v$ , substitute into both equations, then equate real and imaginary parts to get a $4 \times 4$ real system. Solve using elimination or substitution.

Steps:

Use substitution or elimination to reduce to one equation in one unknown.
Separate real and imaginary parts if needed.
Solve the resulting real system.

Exercise 1.4 — Question 4

This exercise covers the equality of complex numbers, basic algebraic operations, and solving simultaneous linear equations with complex coefficients.

Key Concepts

1. Equality of Complex Numbers (M-11-A-03)

Two complex numbers $z_{1} = a + ib$ and $z_{2} = c + i d$ are equal if and only if: $a = c and b = d$ That is, their real parts are equal and their imaginary parts are equal.

Example: If $(2 x - 1) + (y + 3) i = 5 + 2 i$ , then:

Real parts: $2 x - 1 = 5 \Rightarrow x = 3$
Imaginary parts: $y + 3 = 2 \Rightarrow y = - 1$

2. Basic Operations on Complex Numbers (M-11-A-04)

For $z_{1} = a + ib$ and $z_{2} = c + i d$ :

Operation	Result
Addition	$(a + c) + i (b + d)$
Subtraction	$(a - c) + i (b - d)$
Multiplication	$(a c - b d) + i (a d + b c)$
Division	$\frac{( a c + b d ) + i ( b c - a d )}{c ^{2} + d ^{2}}$

Modulus of a product: $∣ z_{1} z_{2} \dots z_{n} ∣ = ∣ z_{1} ∣∣ z_{2} ∣ \dots ∣ z_{n} ∣$

Squaring both sides, if $(x_{1} + i y_{1}) (x_{2} + i y_{2}) \dots (x_{n} + i y_{n}) = a + ib$ , then: $(x_{1}^{2} + y_{1}^{2}) (x_{2}^{2} + y_{2}^{2}) \dots (x_{n}^{2} + y_{n}^{2}) = a^{2} + b^{2}$

Example: $(1 + 2 i) (3 - i) = 3 - i + 6 i - 2 i^{2} = 3 + 5 i + 2 = 5 + 5 i$ .

Check using modulus: $(1^{2} + 2^{2}) (3^{2} + 1^{2}) = 5 \times 10 = 50 = 5^{2} + 5^{2}$ ✓

3. Simultaneous Linear Equations with Complex Coefficients (M-11-A-07)

To solve a system of simultaneous equations with complex coefficients, use elimination or substitution.

Example: Solve $z + i w = 1 + 2 i \dots (1)$ and $i z - w = 3 \dots (2)$

Method — Elimination:

Multiply equation (1) by $i$ : $i z + i^{2} w = i (1 + 2 i) \Rightarrow i z - w = i + 2 i^{2} = - 2 + i \dots (3)$

Subtract equation (2) from (3): $(i z - w) - (i z - w) = (- 2 + i) - 3$ $0 = - 5 + i (contradiction — system is inconsistent for these values)$

Note: Not every system has a solution. If elimination leads to a contradiction (e.g. $0 = k$ where $k \neq = 0$ ), the system is inconsistent (no solution).

Example of a consistent system: Solve $z + i w = 2 + i \dots (1)$ and $i z + w = 1 + 3 i \dots (2)$

Substitute back: $z = 2 + i - i (1 + \frac{i}{2}) = 2 + i - i + \frac{1}{2} = \frac{5}{2}$

General Method (substitution of $z = x + i y$ ):

Steps:

Use substitution or elimination to reduce to one equation in one unknown.
Separate real and imaginary parts if needed.
Solve the resulting real system.