代数式:

$(a+bi) + (c+di) = (a + c) + (b + d)i$

$(a+bi) - (c+di) = (a - c) + (b - d)i$

$(a+bi)(c+di) = (ac-bd)+(ad+bc)i$

$\frac{a+bi}{c+di} = \frac{(a+bi)(c-di)}{(c+di)(c-di)} = \frac{ac+bd}{c^2+d^2} + \frac{bc-ad}{c^2+d^2}i$ (c,d 不同时为0)

三角式：

$r_1(cos\theta_1 + isin\theta_1) \cdot r_2(cos\theta_2 + isin\theta_2) = r_1r_2[cos(\theta_1+\theta_2) +isin(\theta_1+\theta_2)]$

$\frac{r_1(cos\theta_1+isin\theta_1)}{r_2(cos\theta_2+isin\theta_2)} = \frac{r_1}{r_2}[cos(\theta_1 - \theta_2) +isin(\theta_1 - \theta_2)]$

$[r(cos\theta + isin\theta)]^n = r^n \cdot (cosn\theta + isinn\theta)$

$r(cos\theta+isin\theta)$的$n$次方根是：

$\sqrt[n]{r}(cos\frac{\theta+2k\pi}{n} + isin\frac{\theta+2k\pi}{n}) \quad (k = 1 , 2 ,. . . , n-1)$