Report - Módulo 24 – Números Complexos 2 1 2 ––––– 2 10 2 5 2 · III) Substituindo (I) em (II), temos: [(a – c) + (b – d)i] · [1 + i] = 6 ⇔ ⇔ (a – c) + (a – c)i

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