Exercise 1.4 — Question 3

Problem

Solve the following simultaneous linear equations with complex coefficients:

$z+iw=3+2i$ $iz+w=1+4i$

Solution

Step 1: Label the equations

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

Step 2: Eliminate one variable

Multiply equation (1) by $i$:

$iz+i^{2}w=i(3+2i)$ $iz-w=3i+2i^2$ $iz-w=-2+3i\cdots (3)$

Step 3: Add equations (2) and (3)

$iz+w=1+4i$ $iz-w=-2+3i$

Adding: $2iz=(1+4i)+(-2+3i)$ $2iz=-1+7i$ $z=\frac{-1+7i}{2i}$

Step 4: Simplify $z$

Multiply numerator and denominator by $-i$ (the conjugate of $i$, since $i\cdot (-i)=1$... or multiply by $\frac{-i}{-i}$):

$z=\frac{(-1+7i)(-i)}{2i(-i)}=\frac{i-7i^2}{2}=\frac{i+7}{2}=\frac{7}{2}+\frac{1}{2}i$

$\boxed{z=\frac{7}{2}+\frac{1}{2}i}$

Step 5: Find $w$ using equation (1)

$z+iw=3+2i$ $iw=3+2i-z=3+2i-\frac{7}{2}-\frac{1}{2}i$ $iw=\left(3-\frac{7}{2}\right)+\left(2-\frac{1}{2}\right)i$ $iw=-\frac{1}{2}+\frac{3}{2}i$

Divide both sides by $i$ (multiply by $\frac{-i}{-i}$):

$w=\frac{-\frac{1}{2}+\frac{3}{2}i}{i}=\frac{\left(-\frac{1}{2}+\frac{3}{2}i\right)(-i)}{i(-i)}=\frac{\frac{1}{2}i-\frac{3}{2}{i}^2}{1}=\frac{3}{2}+\frac{1}{2}i$

$\boxed{w=\frac{3}{2}+\frac{1}{2}i}$

Step 6: Verify

Check in equation (2): $iz+w$

$iz=i\left(\frac{7}{2}+\frac{1}{2}i\right)=\frac{7}{2}i+\frac{1}{2}{i}^2=-\frac{1}{2}+\frac{7}{2}i$

$iz+w=\left(-\frac{1}{2}+\frac{7}{2}i\right)+\left(\frac{3}{2}+\frac{1}{2}i\right)=1+4i✓$

Key Concepts Used

• Equality of complex numbers: Two complex numbers are equal if and only if their real parts are equal AND their imaginary parts are equal.
• Basic operations: Addition, subtraction, and multiplication of complex numbers, using $i^2=-1$.
• Solving simultaneous equations: Use elimination or substitution, treating complex numbers as algebraic quantities.
• Dividing by a complex number: Multiply numerator and denominator by the conjugate to simplify.