Se stie ca i = √-1 si ca (a + b)^2 = a^2 +2ab + b^2.
(1 + i)^10 + (1 - i)^10 = ((1 + √-1)^2)^5 + ((1 - √-1)^2)^5 = (1^2 + 2√-1 + (√-1)^2)^5 + (1^2 - 2√-1 + (√-1)^2)^5 = (1 - 1 + 2√-1)^5 + (1 - 1 - 2√-1)^5 = (2√-1)^5 + (- 2√-1)^5 = 32*(-1)*√-1 - 32*(-1)*√-1 = 0.
Raspuns: (1 + i)^10 + (1 - i)^10 = 0.
Sper ca te-am ajutat!
