复数

$|z|=|a+b\mathrm{i}|=\sqrt{a^2+b^2}$ $(a+b\mathrm{i})\pm (c+d\mathrm{i})=(a\pm c)+(b\pm d)\mathrm{i}\\(a+b\mathrm{i})(c+d\mathrm{i})=(ac-bd)+(ad+bc)\mathrm{i}\\\frac{a+b\mathrm{i} }{c+d\mathrm{i} }=\frac{(a+b\mathrm{i})(c-d\mathrm{i})}{(c+d\mathrm{i})(c-d\mathrm{i})}=\frac{(ac+bd)+(bc-ad)\mathrm{i} }{c^2+d^2}$ $z = a + b\text{i} = r(\cos \theta + \text{isin}\theta ) = re^{\text{i}\theta }$