第11章 11.2

(1)

$(2+3i)(4-i) = 8 + 10i + 3 = 11 + 10i$

(2)

$(2-7i)^2 = 4 - 28i - 49 = -45 - 28i$

(3)

$\frac{3+2i}{2-i} = \frac{(3+2i)(2+i)}{4+1} = \frac{1}{5}(6+7i-2) = \frac{4}{5} + \frac{7}{5}i$

(4)

$\frac{3-i}{1+2i} + \frac{5+2i}{3-i} = \frac{(3-i)(1-2i)}{1+4} + \frac{(5+2i)(3+i)}{9+1} = \frac{1}{5}(3-7i-2) + \frac{1}{10}(15+11i-2)$

$= \frac{1}{5} - \frac{7}{5}i + \frac{13}{10} + \frac{11}{10}i = \frac{3}{2} - \frac{3}{10}i$

(5)

$i^5 + \frac{1}{i} = i - i = 0$

(6)

$(2-i)(1+i)(1-2i) = (3+i)(1-2i) = 3 - 5i + 2 = 5 - 5i$