Thực hiện các phép tính sau:

a) $(3+2\mathrm{i})\left[\right(2-\mathrm{i})+(3-2\mathrm{i}\left)\right]$;            b) (4 - 3i) + $\frac{1+i}{2+i}$;

c) (1 +i)² - (1 - i)²;                        d) $\frac{3+i}{2+i}-\frac{4-3i}{2-i}.$

Giải:

a) $(3+2\mathrm{i})\left[\right(2-\mathrm{i})+(3-2\mathrm{i}\left)\right]$ = (3 + 2i)(5 - 3i) = 21 + i

b) (4 - 3i) + $\frac{1+\mathrm{i}}{2+\mathrm{i}}$ = (4 - 3i) + $\frac{(1+\mathrm{i})(2-\mathrm{i})}{5}$=(4 - 3i) + $\left(\frac{3}{5}+\frac{1}{5}\mathrm{i}\right)$

                           = $\left(4+\frac{3}{5}\right)-\left(3-\frac{1}{5}\right)\mathrm{i}=\frac{23}{5}-\frac{14}{5}\mathrm{i}$

c) (1 + i)² - (1- i)² = 2i - (-2i)=4i

d) $\frac{3+\mathrm{i}}{2+\mathrm{i}} - \frac{4 - 3\mathrm{i}}{2 - \mathrm{i}} = \frac{(3 + \mathrm{i})(2 - \mathrm{i})}{5}-\frac{(4 -3\mathrm{i})(2-\mathrm{i})}{5}=\frac{7-\mathrm{i}}{5}-\frac{11-2\mathrm{i}}{5}=-\frac{4}{5}+\frac{1}{5}\mathrm{i}.$