Resolu\u00e7\u00e3o<\/b><\/span><\/p>\nOra, $$\\begin{array}{*{20}{l}}
\n{{{\\left( {\\sqrt 3\u00a0 – i} \\right)}^k}}& = &{{{\\left[ {2\\left( {\\frac{{\\sqrt 3 }}{2} – \\frac{1}{2}i} \\right)} \\right]}^k}} \\\\
\n{}& = &{{{\\left[ {2\\operatorname{cis} \\left( {\\frac{{11\\pi }}{6}} \\right)} \\right]}^k}} \\\\
\n{}& = &{{2^k}\\operatorname{cis} \\left( {\\frac{{11k\\pi }}{6}} \\right)}
\n\\end{array}$$<\/p>\n
Para que ${{{\\left( {\\sqrt 3\u00a0 – i} \\right)}^k}}$ represente um n\u00famero real positivo, ter\u00e1 de ser:<\/p>\n
$$\\frac{{11k\\pi }}{6} = 2p\\pi ,p \\in \\mathbb{Z}$$<\/p>\n
ou seja, $$k = \\frac{{12p}}{{11}},p \\in \\mathbb{Z}$$<\/p>\n
Portanto, o menor n\u00famero inteiro positivo que satisfaz a condi\u00e7\u00e3o \u00e9 $k = 12$, obtido com $p = 11$.
\n\u00ad<\/p>\n