Exercise 1.3 — Question 2

Topic: Simultaneous Linear Equations with Complex Coefficients

SLO: Solve simultaneous linear equations with complex coefficients.

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Key Method

To solve a system of simultaneous linear equations where the coefficients are complex numbers, use elimination or substitution, treating $i$ as an algebraic constant with $i^2=-1$.

Steps:

1. Label the equations (1) and (2).
2. Eliminate one unknown by adding/subtracting suitable multiples of the equations.
3. Solve for the remaining unknown by dividing by its complex coefficient (multiply numerator and denominator by the conjugate).
4. Back-substitute to find the other unknown.
5. Verify by substituting both values into the original equations.

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Worked Example

Solve the simultaneous equations:

$z+iw=2\cdots (1)$ $iz-w=3\cdots (2)$

Step 1: Multiply equation (1) by $i$:

$iz+i^{2}w=2i$ $iz-w=2i\cdots (3)[\text{since }{i}^2=-1]$

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

$(iz-w)-(iz-w)\Rightarrow$ wait — add (3) and (2):

From (3): $iz-w=2i$

From (2): $iz-w=3$

Actually, multiply (1) by $i$: $iz-w=2i\cdots (3)$

Subtract (2) from (3): $(iz-w)-(iz-w)=2i-3$

This gives $0=2i-3$, which is a contradiction — so let us use a different elimination.

Correct approach — eliminate $w$:

From (1): $z+iw=2\Rightarrow z=2-iw\cdots (3)$

Substitute into (2): $i(2-iw)-w=3$ $2i-i^{2}w-w=3$ 2i + w - w = 3 \quad [\text{since } -i^2 = 1] $2i=3$

This system has no solution unless the equations are set up differently. The method below applies to a general consistent system.

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General Worked Example (Consistent System)

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

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

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

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

Expand $(1+i)i=i+i^2=i-1=-1+i$

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

Step 3: Solve for $w$: w = \frac{4-2i}{2-i}

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

Step 4: Substitute $w=2$ into (3): $z=1+2i-i(2)=1+2i-2i=1$

Solution: $z=1,w=2$

Verification:

• Eq (1): $1+i(2)=1+2i$ ✓
• Eq (2): $(1+i)(1)+2=1+i+2=3+i$ ✓

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Key Formula: Division by a Complex Number

To divide by a complex number $c+di$, multiply numerator and denominator by its conjugate $c-di$:

$\frac{a+bi}{c+di} = \frac{(a+bi)(c-di)}{c^2+d^2}$$

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