Exercise 1.4 — Q1: Simultaneous Linear Equations with Complex Coefficients

Key Method

To solve simultaneous linear equations with complex coefficients, treat the equations algebraically just as you would real-coefficient systems, but apply complex arithmetic throughout.

General approach:

1. Write the system in the form: $a_{1}z+b_{1}w=c_1$ $a_{2}z+b_{2}w=c_2$ where $a_1,b_1,c_1,a_2,b_2,c_2\in \mathbb{C}$.

2. Use elimination or substitution to reduce to one equation in one unknown.

3. Solve for the unknown complex number, then back-substitute.

4. Verify by substituting both values into the original equations.

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Worked Example

Solve the system: $z+iw=2+3i\cdots (1)$ $(1+i)z-w=3-i\cdots (2)$

Step 1: From equation (1), express $z$: $z=2+3i-iw$

Step 2: Substitute into equation (2): $(1+i)(2+3i-iw)-w=3-i$

Expand $(1+i)(2+3i)$: $(1+i)(2+3i)=2+3i+2i+3i^2=2+5i-3=-1+5i$

Expand $(1+i)(-iw)$: $(1+i)(-iw)=(-i-i^2)w=(-i+1)w=(1-i)w$

So equation (2) becomes: $(-1+5i)+(1-i)w-w=3-i$ $(1-i-1)w=3-i+1-5i$ $-iw=4-6i$

Step 3: Solve for $w$: $w=\frac{4-6i}{-i}=\frac{(4-6i)}{-i}\cdot \frac{i}{i}=\frac{(4-6i)i}{-i^2}=\frac{4i-6i^2}{1}=6+4i$

Step 4: Back-substitute to find $z$: $z=2+3i-i(6+4i)=2+3i-6i-4i^2=2+3i-6i+4=6-3i$

Solution: $z=6-3i$, $w=6+4i$

Verification in (1): $(6-3i)+i(6+4i)=6-3i+6i+4i^2=6+3i-4=2+3i$ ✓

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Key Reminders

• $i^2=-1$ — apply this whenever expanding.
• To divide by a complex number $a+ib$, multiply numerator and denominator by its conjugate $a-ib$.
• Two complex numbers are equal iff their real parts are equal and their imaginary parts are equal.