Question

пожалуйста, решение!
1-3i\1+3i-1+3i\1-3i=?

Answer 1

Ответ:

$\dfrac{1-3i}{1+3i}-\dfrac{1+3i}{1-3i}=\dfrac{(1-3i)^2-(1+3i)^2}{(1-3i)(1+3i)}=\dfrac{(1-6i+9i^2)-(1+6i+9i^2)}{1-9i^2}=\\\\\\=\dfrac{-12i}{1+9}=-1,2\, i$