Question

putin ajutor aici? Mulțumiesccc anticipat

Answer 1

Numerele acestea au o proprietate interesanta daca le ridici la patrat:

(1 + i)² = 1 + 2i + i² = 1 + 2i - 1 = 2i
(1 - i)² = 1 - 2i - 1 = -2i

Astfel, vom scrie cei doi termeni in functie de (1+i)² si (1-i)²

[tex]z= \frac{(1+i)^{2*1007}}{(1-i)^{2*1006+1}} = \frac{((1+i)^2)^{1007}}{((1-i)^2)^{1006}*(1-i)}= \frac{(2i)^{1007}}{(-2i)^{1006}*(1-i)}= \frac{2i}{1-i} = \frac{2i(1+i)}{(1-i)(1+i)} \\ z= \frac{2i-2}{2}=i-1 [/tex]

Modulul: |z| = √(1² + (-1)²) = √2