Question

(1+i)^2008 + (1-i)^2008 nu imi dau seama cum se rezolva.Initial m-am gandit la binomul lui Newton dar nu mi-a iesit nimic.

Answer 1

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Answer 2

(1+i)²⁰⁰⁸+(1-i)²⁰⁰⁸=[(1+i)²]¹⁰⁰⁴+[(1-i)²]¹⁰⁰⁴=(1+i²+2i)¹⁰⁰⁴+(1+i²-2i)¹⁰⁰⁴=(1-1+2i)¹⁰⁰⁴+(1-1-2i)¹⁰⁰⁴=(2i)¹⁰⁰⁴+(-2i)¹⁰⁰⁴=2¹⁰⁰⁴+2¹⁰⁰⁴=2¹⁰⁰⁴(1+1)=2¹⁰⁰⁴*2=2¹⁰⁰⁴*2¹=2¹⁰⁰⁴⁺¹=2¹⁰⁰⁵

Aplici puterile lui i : i²=-1;i³=-i;i⁴=1 ; i⁵=i, .... i¹⁰⁰³=-i;i¹⁰⁰⁴=1, etc