Question

Тема: Комплексные числа
Помогите люди добрые.
С 1 по 5​

Answer 1

Ответ:

$\boxed{\ i^2=-1\ }\\\\1)\ \ (-5+2i)+(5+2i)=(-5+5)+(2i+2i)=4i\\\\2)\ \ (-3-5i)+(7-2i)=(-3+7)+(-5i-2i)=4-7i\\\\3)\ \ (3+2i)(1+i)=3+3i+2i+2i^2=3+5i-2=1+5i\\\\4)\ \ (6+4i)\cdot 3i=18i+12i^2=-12+18i\\\\5)\ \ (2-3i)\cdot (-5i)=-10i+15i^2=-15-10i\\\\6)\ \ (1-5i)^2=1^2-2\cdot 5i+(5i)^2=1-10i+25i^2=1-10i-25=\\\\=-24-10i\\\\7)\ \ (5-i)^3=5^3-3\cdot 25i+3\cdot 5i^2-i^3=125-75i-15-(i^2\cdot i)=\\\\=110-75i-(-i)=110-75i+i=110-74i$

$8)\ \ \dfrac{1-3i}{1+3i}=\dfrac{(1-3i)^2}{(1+3i)(1-3i)}=\dfrac{1-6i+9i^2}{1-9i^2}=\dfrac{1-6i-9}{1+9}=\dfrac{-8-6i}{10}=\\\\=-\dfrac{8}{10}-\dfrac{6i}{10}=-\dfrac{4}{5}-\dfrac{3}{5}\, i$