Question

aratati ca N=(4+3i)^2 + (3-4i)^2 este natural , i^2 = -1

Multumesc!

Answer 1

N = (4+3i)² + (3-4i)²⏐· i²

Ni² = i²(4+3i)² + i²(3-4i)²

Ni² = (4i+3i²)² + (3i-4i²)²

-N = (-3+4i)² + (4+3i)²

⇒ N = -N

⇒ N+N = 0

⇒ 2N = 0

⇒ N = 0 ∈ ℕ

Answer 2

$N=(4+3i)^2+(3-4i)^2\\ \\ N=4^2+2\cdot 4 \cdot 3i +3^2\cdot i^2 + 3^2 - 2 \cdot 3 \cdot 4i + 4^2 \cdot i^2\\ \\ N=16+24i+9\cdot (-1)+9-24i+16\cdot (-1)\\ \\ N=16-9+9-16\\ \\ N=0 \in \mathbb{N}$