[tex]i\cdot i^2\cdot i^3\cdot ...\cdot i^{2013} = \\ \\=i^1\cdot i^2\cdot i^3\cdot ...\cdot i^{2013}\\ \\ = i^{1+2+3+...+2013} = \\ \\ = i^{\dfrac{2013\cdot (2013+1)}{2}} = \\ \\ =i^\dfrac{2013\cdot 2014}{2}} = \\ \\ =i^\big{2013\cdot 1007} [/tex]
Avem un ciclu:
[tex]i^1 = i\\ i^2 = -1 \\ i^3 = -i \\ i^4 = 1 \\ ------ \\ i^5 = i \\ i^6 = -1 \\ i^6 = -i \\ ......[/tex]
Avem niste conditii, pentru i^n;
1) Daca n impar si:
a) Daca (n+1)/2 impar => i^n = i
b) Daca (n+1)/2 par => i^n = -i
2) Daca n par si:
a) Daca n/2 impar => i^n = -1
b) Daca n/2 par => i^n = 1
Noi avem:
i^{2013×1007}
U2 (2013×1007) = U2 (13×7) = 91 (U2 -> ultimele doua cifre)
91 impar:
(91+1)/2 = 46 par => i^(2013×1007) = -i
[tex]\Rightarrow \boxed{i\cdot i^2\cdot i^3\cdot ...\cdot i^{2013} = -i}[/tex]
