Question

помогите пожалуйста по математике ​

Answer 1

Ответ:

$z_1=2-3i\ \ ,\ \ z_2=5+2i\ \ \ \ \ \ \boxed{\ i^2=-1\ }\\\\\\z_1+z_2=(2-3i)+(5+2i)=(2+5)+(-3i+2i)=7-i\\\\z_1-z_2=(2-3i)-(5+2i)=(2-5)+(-3i-2i)=-3-5i\\\\z_1\cdot z_2=(2-3i)(5+2i)=10+4i-15i-6i^2=10-11i+6=16-11i\\\\\\\dfrac{z_1}{z_2}=\dfrac{2-3i}{5+2i}=\dfrac{(2-3i)(5-2i)}{(5+2i)(5-2i)}=\dfrac{10-4i-15i+6i^2}{25-4i^2}=\dfrac{10-19i-6}{25+4}=\\\\\\=\dfrac{4-19i}{29}=\dfrac{4}{29}-\dfrac{19}{29}\, i$