Даны комплексные числа z1=2-5i и z2=6-8i Найти: a) z1+z2; б) z1-z2; в)z1 z2; г) z1/z2

Решите задачу:

$z_1=2-5i \\z_2=6-8i \\ \\ z_1+z_2=2-5i+6-8i=8-13i \\ z_1-z_2=2-5i-(6-8i)=2-5i-6+8i=-4+3i \\ z_1\cdot z_2=(2-5i)(6-8i)=12-16i-30i+40i^2=-28-46i \\$ $\frac{z_1}{z_2}= \frac{2-5i}{6-8i}= \frac{(2-5i)(6+8i)}{(6-8i)(6+8i)} = \frac{12+16i-30i-40i^2}{6^2-64i^2} = \frac{12+40-14i}{36+64}= \frac{52-14i}{100} =$ $= \frac{2(26-7i)}{50\cdot2} = \frac{26-7i}{50} = \frac{26}{50} - \frac{7i}{50} = \frac{13}{25} - \frac{7i}{50.}$