Matemática, perguntado por CarlosIMEITA, 10 meses atrás

ITA - Números Complexos (Segue imagem anexa)

Anexos:

Soluções para a tarefa

Respondido por elizeugatao
2

\displaystyle \sf I) \ (cos\ \alpha +isen\ \alpha)^{10} = cos \ 10\alpha +i\sen \ 10\alpha \  \ \huge{(\sf VERDADEIRA)}}\checkmark

\displaystyle \underline{\sf coment{\'a}rio} : \\\\ \boxed{\begin{array}{I}\sf \underline{Primeira\ Formula \ de \ Moivre}\  \\\\  \sf Z^{n}=|Z|^{n}\cdot\left(cos \ \alpha \ +i.\ sen\alpha\right)^{n} \\\\ \sf \underline{Da{\'i}}: \\\\ \sf \sf \left(cos \ \alpha \ +i.\ sen \alpha\right)^{10}=cos\ 10\alpha+i.sen\ 10\alpha   \end{array}} \checkmark

2)

\boxed{\begin{array}{I} \displaystyle \sf II)\ \frac{5i}{2+i}= 1+2i \\\\\\ \underline{\text{Racionalizando o lado esquerdo}}: \\\\ \displaystyle \sf \frac{5i}{(2+i)}\frac{(2-i)}{(2-i)}=1+2i\ \  \to \ \  \frac{10i-5i^2}{4-i^2} = 1+2i  \\\\\\ \displaystyle \sf \frac{10i-5(-1)}{4-(-1)} = 1+2i \ \ \ \ \to \ \ \frac{10i+5}{5}=1+2i \\\\\\  \boxed{\sf 1+2i=1+2i } \checkmark \\ \  \end{array}} \text{(VERDADEIRO)}

3)

\boxed{\begin{array}{I} \sf (1-i)^4 = - 4 \\\\ \sf\underline{\text{Desenvolvendo o lado esquerdo}}: \\\\ \left[\ \sf (1-i)^2\ \right]^2 =-4 \\\\ \sf \left[1-2i+i^2\right]^2 = - 4 \\\\ \sf \left[ 1 -2i -1 \right]^2=-4 \\\\ \sf (-2i)^2= -4 \\\\ \sf 4i^2 = - 4 \\\\ \boxed{-4 = -4}\checkmark \\ \   \end{array}} \text{(VERDADEIRO)}

4)

\boxed{\begin{array}{I} \sf Z^2 = \left |\bar{Z}\right |^2 \ , \text{ent{\~a}o Z {\'e} real ou imagin{\'a}rio puro }\\\\ \underline{\text{Fa{\c c}amos Z = a+b.i }}: \\\\ \sf \left(a+b.i\right)^2 = \left (\sqrt{a^2+b^2}\right)^2 \\\\ \sf a^2+2a.b.i+b^2i^2=a^2+b^2\\\\ \sf a^2-b^2+2a.b.i  =a^2+b^2 \\\\ \sf \underline{Da{\'i}}: \\\\ \sf a^2-b^2=a^2+b^2 \to 2b^2 = 0 \to \boxed{\sf b = 0 }\\\\ \sf 2a.b.i = 0 \to \boxed{\sf a = b = 0}  \end{array}} \ \text{(VERDADEIRO)}

5)

\boxed{\begin{array}{I} \sf x^4+x^3-x-1 \ \text{possui apenas ra{\'i}zes reais}\\\\ \underline{\sf Fatorando}:\\\\ \sf x^3(x+1) -(x+1)\\\\ \sf (x+1)(x^3+1 )\\\\ \sf(x+1)(x+1)(x^2-x+1) \\\\ \sf \underline{ra{\'i}zes}: \\\\ \boxed{\sf x = -1 }\\\\ \sf x^2-x+1 \to \displaystyle \boxed{\sf x= \frac{1\pm i\cdot\sqrt{3}}{2}}\\ \ \end{array}} \text{(FALSO)}

APENAS QUATRO VERDADEIRAS (B)

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