1.4 Q-7

Exercise 1.4 — Question 7

Problem

Solve the following simultaneous linear equations with complex coefficients:

$z+(1+i)w=3+2i$ $(2-i)z+w=4-i$

Method: Elimination

We use elimination to solve for $z$ and $w$.

Step 1: Label the equations.

$z+(1+i)w=3+2i\cdots (1)$ $(2-i)z+w=4-i\cdots (2)$

Step 2: Eliminate one variable.

Multiply equation (1) by $(2-i)$:

$(2-i)z+(2-i)(1+i)w=(2-i)(3+2i)$

Compute $(2-i)(1+i)$: $(2-i)(1+i)=2+2i-i-i^2=2+i+1=3+i$

Compute $(2-i)(3+2i)$: $(2-i)(3+2i)=6+4i-3i-2i^2=6+i+2=8+i$

So the modified equation (1) becomes: $(2-i)z+(3+i)w=8+i\cdots (3)$

Step 3: Subtract equation (2) from equation (3).

$[(2-i)z+(3+i)w]-[(2-i)z+w]=(8+i)-(4-i)$

$(3+i-1)w=4+2i$

$(2+i)w=4+2i$

Step 4: Solve for $w$.

$w=\frac{4+2i}{2+i}$

Multiply numerator and denominator by the conjugate of the denominator $(2-i)$:

$w=\frac{(4+2i)(2-i)}{(2+i)(2-i)}=\frac{8-4i+4i-2i^2}{4-i^2}=\frac{8+2}{4+1}=\frac{10}{5}=2$

So $w=2$.

Step 5: Substitute $w=2$ into equation (1) to find $z$.

$z+(1+i)(2)=3+2i$

$z+2+2i=3+2i$

$z=3+2i-2-2i=1$

So $z=1$.

Solution

$\boxed{z=1,w=2}$

Verification

Check in equation (1): $1+(1+i)(2)=1+2+2i=3+2i$ ✓

Check in equation (2): $(2-i)(1)+2=2-i+2=4-i$ ✓

Key Concepts Used

• Basic operations on complex numbers (M-11-A-04): multiplication, addition, and division of complex numbers including rationalising by multiplying by the conjugate.
• Solving simultaneous equations with complex coefficients (M-11-A-07): elimination method applied to a $2\times 2$ system.