а) 1/i = i/i^2 = i/-1 = -i
б) 1/(1+i) = (1-i)/(1+i)(1-i) = (1-i)/(1-i^2) = (1-i)/2 = 1/2 - 1/2i
в) (5-i)/(i+2) = [(5-i)(i-2)]/(i+2)(i-2) = (5i - 10 - i^2 + 2i)/(i^2 - 2i + 2i - 4)= (6 + 7i)/5 = 6/5 + 7/5i.
а) 1/i = i/i^2 = i/-1 = -i
б) 1/(1+i) = (1-i)/(1+i)(1-i) = (1-i)/(1-i^2) = (1-i)/2 = 1/2 - 1/2i
в) (5-i)/(i+2) = [(5-i)(i-2)]/(i+2)(i-2) = (5i - 10 - i^2 + 2i)/(i^2 - 2i + 2i - 4)
= (6 + 7i)/5 = 6/5 + 7/5i.