Đáp án: $z^2+\overline{z}^2=-\dfrac{416}{25}$
Giải thích các bước giải:
Ta có:
$z=2+3i-\dfrac{(5+2i)(3+2i)}{3-i}$
$\to z=2+3i-\dfrac{15+16i+4i^2}{3-i}$
$\to z=2+3i-\dfrac{15+16i-4}{3-i}$
$\to z=2+3i-\dfrac{11+16i}{3-i}$
$\to z=2+3i-\dfrac{(11+16i)(3+i)}{(3-i)(3+i)}$
$\to z=2+3i-\dfrac{17+59i}{3^2-i^2}$
$\to z=2+3i-\dfrac{17+59i}{3^2+1}$
$\to z=2+3i-\dfrac{17+59i}{10}$
$\to z=\dfrac{3}{10}-\dfrac{29}{10}i$
$\to \overline{z}=\dfrac{3}{10}+\dfrac{29}{10}i$
$\to z^2+\overline{z}^2=(\dfrac{3}{10}-\dfrac{29}{10}i)^2+(\dfrac{3}{10}+\dfrac{29}{10}i)^2$
$\to z^2+\overline{z}^2=(\dfrac{3}{10})^2-2\cdot \dfrac3{10}\cdot \dfrac{29}{10}i+(\dfrac{29}{10}i)^2+(\dfrac{3}{10})^2+2\cdot \dfrac3{10}\cdot \dfrac{29}{10}i+(\dfrac{29}{10}i)^2$
$\to z^2+\overline{z}^2=2((\dfrac{3}{10})^2+(\dfrac{29}{10}i)^2)$
$\to z^2+\overline{z}^2=2(\dfrac{9}{100}-\dfrac{29^2}{100})$
$\to z^2+\overline{z}^2=-\dfrac{416}{25}$
