Bài 1:
a) $A =\dfrac{1}{(1-i)(4-3i)}$
$\to A = \dfrac{(1+i)(4+3i)}{(1-i^2)(16 - 9i^2)}$
$\to A =\dfrac{4 + 7i + 3i^2}{(1+1)(16+9)}$
$\to A = \dfrac{4+ 7i - 3}{50}$
$\to A = \dfrac{1}{50} + \dfrac{7}{50}$
b) $B = (3-2i)i$
$\to B = 3i - 2i^2$
$\to B = 2 + 3i$
c) $C = (1- i^2)$
$\to C = 1- 2i + i^2$
$\to C = 1 - 2i - 1$
$\to C = - 2i$
d) $D = \dfrac{3-2i}{i}$
$\to D = \dfrac{(3-2i)i}{i^2}$
$\to D =\dfrac{2 + 3i}{-1}$
$\to D = - 2 - 3i$
Bài 2:
$\quad z =\dfrac{- 5 + 6i}{4 + 3i}$
$\to z =\dfrac{(-5 +6i)(4 - 3i)}{16 - 9i^2}$
$\to z =\dfrac{-20 + 29i - 18i^2}{16 + 9}$
$\to z =\dfrac{- 20 + 29i + 18}{25}$
$\to z =- \dfrac{2}{25} +\dfrac{29}{25}i$
Ta được:
$|z|=\sqrt{\left(-\dfrac{2}{25}\right)^2 + \left(\dfrac{29}{25}\right)^2}= \dfrac{13\sqrt5}{25}$
$\overline{z}= -\dfrac{2}{25} - \dfrac{29}{25}i$
