Question

2.Sa se calculeze(1+i)²⁰.​

Answer 1

$\displaystyle{ (1 + i)^{20} = [(1 + i)^{2}]^{10} }$

$\displaystyle{ (1+i)^{2} = 1^{2} + 2i + i^{2} = 1 + 2i - 1 = 2i }$

$\displaystyle{ \rightarrow (2i)^{10} = 2^{10} \cdot i^{10}}$

$\displaystyle{ i^{10} = i^{8 + 2} = i^{8} \cdot i^{2} = i^{2} = -1 }$

$\displaystyle{ \rightarrow (1 + i)^{20} = 2^{10} \cdot -1 = -2^{10} = -1024 }$