3.3 Multiplication
Complex numbers are multiplied by expanding the brackets in the ordinary way and then replacing \(i^{2}\) by \(-1\). Doing that once in general gives a formula worth recording.
Definition 3.3 (Product). Let \(z_1=x_1+iy_1\) and \(z_2=x_2+iy_2\). Then \[z_1z_2=\left (x_1x_2-y_1y_2\right )+i\left (x_1y_2+x_2y_1\right ).\]
Note 3.4. The formula is worth seeing once and then forgetting: expanding \(\left (x_1+iy_1\right )\left (x_2+iy_2\right )\) directly and using \(i^{2}=-1\) is quicker and harder to get wrong than remembering which product carries the minus sign. The minus appears because \(iy_1\cdot iy_2=i^{2}y_1y_2=-y_1y_2\), and that is the only place \(i^{2}\) arises.
Solution. (a). By the product formula, \begin {align*} (3+2i)(4-5i) &= (3\times 4-(-5\times 2))+i(3(-5)+4(2))\\ &= (12+10)+i(-15+8)\\ &= 22-7i . \end {align*}
Expanding directly gives the same answer and is usually quicker than quoting the formula: \(12-15i+8i-10i^{2}=12-7i+10=22-7i\), using \(i^{2}=-1\).
(b). The second factor is the conjugate of the first, so the imaginary parts cancel: \[\left (3+\sqrt {2}i\right )\left (3-\sqrt {2}i\right )=3^{2}-\left (\sqrt {2}i\right )^{2} =9-2i^{2}=9+2=11 .\] The answer is real, as it had to be — a number times its conjugate always gives \(x^{2}+y^{2}\), which is \(\left |z\right |^{2}\).
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