Exercise 1.4 — Question 2: Simultaneous Linear Equations with Complex Coefficients

This exercise covers solving systems of simultaneous linear equations where the coefficients and/or constants are complex numbers.

Key Method

To solve simultaneous equations with complex coefficients, use the equality of complex numbers: if $a+ib=c+id$, then $a=c$ (real parts equal) and $b=d$ (imaginary parts equal).

Steps:

1. Write the system of equations.
2. Use elimination or substitution to isolate one unknown.
3. Separate real and imaginary parts to find the solution.

---

Worked Example 1

Solve for $z_1$ and $z_2$: $z_1+z_2=3+2i$ $z_1-z_2=1+4i$

Solution:

Adding both equations: $2z_1=(3+2i)+(1+4i)=4+6i$ $z_1=2+3i$

Subtracting the second from the first: $2z_2=(3+2i)-(1+4i)=2-2i$ $z_2=1-i$

Verification: $z_1+z_2=(2+3i)+(1-i)=3+2i$ ✓ and $z_1-z_2=(2+3i)-(1-i)=1+4i$ ✓

---

Worked Example 2

Solve for $x$ and $y$ (real numbers) given: $(2+i)x+(1-i)y=3+4i$

Solution:

Expand the left side: $(2x+xi)+(y-yi)=3+4i$ $(2x+y)+(x-y)i=3+4i$

Equating real and imaginary parts: $2x+y=3\text{...(i)}$ $x-y=4\text{...(ii)}$

Adding (i) and (ii): $3x=7⟹x=\frac{7}{3}$

From (ii): $y=x-4=\frac{7}{3}-4=-\frac{5}{3}$

---

Worked Example 3

Solve the system: $(1+i)z_1+(2-i)z_2=2+3i$ $(1-i)z_1+(1+i)z_2=1+i$

Solution:

Let $z_1=a+bi$ and $z_2=c+di$ where $a,b,c,d\in \mathbb{R}$.

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

Eq (1) $\times (1-i)$: $(1+i)(1-i)z_1+(2-i)(1-i)z_2=(2+3i)(1-i)$ $2z_1+(2-i-2i+i^2)z_2=2-2i+3i-3i^2$ $2z_1+(1-3i)z_2=5+i\text{...(iii)}$

Eq (2) $\times (1+i)$: $(1-i)(1+i)z_1+(1+i{)}^2{z}_2=(1+i{)}^2$ $2z_1+(1+2i+i^2)z_2=1+2i+i^2$ $2z_1+2i{z}_2=2i\text{...(iv)}$

Subtracting (iv) from (iii): $(1-3i-2i)z_2=5+i-2i$ $(1-5i)z_2=5-i$

$z_2=\frac{5-i}{1-5i}=\frac{(5-i)(1+5i)}{(1-5i)(1+5i)}=\frac{5+25i-i-5i^2}{1+25}=\frac{10+24i}{26}=\frac{5+12i}{13}$

Substitute back into (iv) to find $z_1$.

---

Key Principle

This principle is the foundation for solving simultaneous equations with complex coefficients.