Z1=12+15i Z2=3-4i Найти z1+z2 Z1-z2 Z1/z2 Z1×z2

Решите задачу:

$z_1=12+15i\; \; ,\; \; \; z_2=3-4i\\\\z_1+z_2=12+15i+3-4i=15+11i\\\\z_1-z_2=12+15i-3+4i=9+19i\\\\ \frac{z_1}{z_2}=\frac{(12+15i)(3+4i)}{(3-4i)(3+4i)}=\frac{36+48i+45i+60i^2}{9-16i^2} =[\; i^2=-1\; ]= \frac{36+93i-60}{9+16} = \\\\=\frac{-24+93i}{25}=-\frac{24}{25} +\frac{93}{25}i\\\\z_1\cdot z_2=(12+15i)(3-4i)=36-48i+45i-90i^2=\\\\=36-3i+90=126-3i$