Question

Ma poate ajuta cineva la acest ex va rog mult de tot ​

Answer 1

Răspuns:

Explicație pas cu pas:

Answer 2

 

$\displaystyle\bf\\Folosim~formulele:\\\\(1)~~~(a-b)^2=a^2-2ab+b^2\\\\(2)~~~i^2=-1\\\\Rezolvare:\\\\(1-i)^{24}=\\\\=(1-i)^{2\times12}=\\\\=\Big((1-i)^2\Big)^{12}=\\\\=\Big(1^2-2\times1\times i+i^2\Big)^{12}=\\\\=\Big(1-2i+i^2\Big)^{12}=\\\\=\Big(1-2i-1\Big)^{12}=\\\\=\Big(-2i\Big)^{12}=\\\\=\Big(-2i\Big)^{2\times6}=\\\\=\Big(-2i)^2\Big)^{6}=\\\\=\Big((-2)^2\times(i)^2\Big)^{6}=\\\\=\Big(4\times(-1)\Big)^{6}=\\\\=\Big(-4\Big)^{6}=$

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$\displaystyle\bf\\=\Big(-4\Big)^{2\times3}=\\\\=\Big((-4)^2\Big)^{3}=\\\\=\Big(16\Big)^{3}=\boxed{\bf4096\in N \subset Z}$