In order to add and subtract complex numbers, you just add and subtract the real and imaginary parts separately. That is \[\begin{align*} (a+bi)+(c+di) &= (a+c)+(b+d)i \\ (a+bi)-(c+di) &= (a-c) + (b-d)i. \end{align*}\]

**Worked exercise:** Add the complex numbers \(3+5i\) and \(2-3i\). **Solution:** $$(3+5i)+(2-3i) = (3+2)+(5-3)i = 5 +2i.$$

You multiply complex numbers just like you would multiply binomials, remembering that \(i^2 = -1\).

**Worked exercise:** Evaluate \( (1-i)(2-3i) \).**Solution:** Expanding like you would with binomials, and applying \(i^2 = -1\), we have$$

\begin{align*}

(1-i)(2-3i) &= 2-3i-2i+3i^2 \\

&= (2-3)-(3+2)i \\

&= -1-5i.

\end{align*}

$$

## Problems

- Add the following complex numbers.
- \(14 + 6i\) and \(-1 – i\)
- \(2+6i\) and \(1-i\)
- \(4-i\) and \(-5-9i\)
- \(3+2i\) and \(6i\)

- Subtract the second complex number from the first complex number.
- \(5i\) and \(2-i\)
- \(1-i\) and \(9+7i\)
- \(-7+4i\) and \(2+5i\)
- \(i\) and \(-i\)

- Evaluate the following.
- \((4+i)(-2-3i)\)
- \((-2+4i)(8-i)\)
- \((8-6i)(-1-2i)\)
- \((5+9i)(4-10i)\)