Question

Ex 168… rezolvare completă vă rog

Answer 1

Răspuns:

167)

$\frac{1}{z} = \frac{1}{1+i} = \frac{1-i}{((1+i)(1-i)} = \frac{1-i}{1-i^{2} } = \frac{1-i}{2} }$

168)

$z^{2n} =(z^{2}) ^{n}=(1+i^{2} +2i)^{n}=(1+-1 +2i)^{n}=(2i)^{n}=(2)^{n}(i)^{n}$

zⁿ ∈R ⇒ z²ⁿ = (zⁿ )²∈R, z²ⁿ ≥0 ⇒2ⁿ·iⁿ∈R, ≥0 ⇒iⁿ=1

⇒ $z^{2n} =(2)^{n}$