第4章 4.20

(1)

$(1-2i)^2 = 1-4i-4 = -3-4i$

(2)

$(2-\sqrt5 i)^2 - 4(2-\sqrt5 i) = (2-\sqrt5 i)(2-\sqrt5 i-4) = -(2-\sqrt5 i)(2+\sqrt5 i) = -(4+5) = -9$

(3)

$(-\sqrt{-12})^3 = -(-1)\cdot 12\sqrt{12}i = 12\sqrt{12}i = 24\sqrt3 i$

(4)

$\left(\frac{1+i}{1-i}\right)^3 = \left(\frac{(1+i)^2}{(1-i)(1+i)}\right)^3 = \left(\frac{1+2i-1}{1+1}\right)^3 = i^3 = -i$

(5)

$\frac{1+2i}{4+3i} = \frac{(1+2i)(4-3i)}{(4+3i)(4-3i)} = \frac{4+5i+6}{16+9} = \frac{10+5i}{25} = \frac{2+i}{5}$