Question

(puc-mg) sendo i=✓-1, o valor de (1+i/1-i-2i/1+i)

Answer 1

Na proxima recomendo mais cautela nos parenteses, mas vamos pelo que entendi:

$\dfrac{1+i}{1-i}-\dfrac{2i}{1+i}=\dfrac{1+i}{1-i}\cdot\dfrac{1+i}{1+i}-\dfrac{2i}{1+i}\cdot\dfrac{1-i}{1-i}=\dfrac{(1+i)^2}{1^2-i^2}-\dfrac{2i(1-i)}{1^2-i^2}=\\\\\dfrac{1^2+2i+i^2}{1-(-1)}-\dfrac{2i-2i^2}{1-(-1)}=\dfrac{1+2i-1}{2}-\dfrac{2i+2}{2}=\dfrac{2i-2i-2}{2}=-1$