對於 $i^2=-1$，及任何實數 $a$、$b$、$c$ 及 $d$。
- 複數的加法
$(a+ib)+(c+id)=(a+c)+i(b+d)$
化簡 $(4+5i)+(8-2i)$。$\begin{array}{cl}
& (4+5i)+(8-2i) \\
= & (4+8)+i(5+(-2)) \\
= & 12+3i
\end{array}$ - 複數的減法
$(a+ib)-(c+id)=(a-c)+i(b-d)$
化簡 $(3-2i)-(2+7i)$。$\begin{array}{cl}
& (3-2i)-(2+7i) \\
= & (3-2)+i((-2)-7) \\
= & 1-9i
\end{array}$ - 複數的乘法
$(a+ib)(c+id)=(ac-bd)+i(bc+ad)$
化簡 $(5-2i)(3+9i)$。$\begin{array}{cl}
& (5-2i)(3+9i) \\
= & (5)(3)+(5)(9i)+(-2i)(3)+(-2i)(9i) \\
= & 15+45i-6i-18i^2 \\
= & 15+39i-18(-1) \\
= & 33+39i
\end{array}$ - 複數的除法
$\dfrac{a+ib}{c+id}=\dfrac{(ac+bd)+i(bc-ad)}{c^2+d^2}$
化簡 $\dfrac{2-5i}{2+7i}$。$\begin{array}{cl}
& \dfrac{2-5i}{2+7i} \\
= & \dfrac{2-5i}{2+7i} \times \dfrac{2-7i}{2-7i} \\
= & \dfrac{(2-5i)(2-7i)}{(2+7i)(2-7i)} \\
= & \dfrac{(2)(2)+(2)(-7i)+(-5i)(2)+(-5i)(-7i)}{2^2-(7i)^2} \\
= & \dfrac{4-14i-10i+35i^2}{4-49i^2} \\
= & \dfrac{4-24i+35(-1)}{4-49(-1)} \\
= & \dfrac{-31-24i}{53} \\
= & \dfrac{-31}{53}-\dfrac{24}{53}i
\end{array}$
