Matemática
ayresrafael10
3

a pergunta e resposta da 10 e11

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(1) Respostas
Jailson123

a) [latex]\left(\dfrac{1}{2}+2i\right)\left(\dfrac{1}{3}-3i\right)=\dfrac{1}{2}\cdot\dfrac{1}{3}-\dfrac{3i}{2}+\dfrac{2i}{3}-6i^2[/latex] [latex]i^2=-1[/latex] e [latex]\text{mmc}(2,3)=6[/latex] [latex]\left(\dfrac{1}{2}+2i\right)\left(\dfrac{1}{3}-3i\right)=\dfrac{1}{6}-\dfrac{3i}{2}+\dfrac{2i}{3}-6\cdot(-1)[/latex] [latex]\left(\dfrac{1}{2}+2i\right)\left(\dfrac{1}{3}-3i\right)=\dfrac{1-9i+6i+36}{6}[/latex] b) [latex]\left(\dfrac{1}{2}+2i\right)\left(\dfrac{1}{3}-3i\right)=\dfrac{37}{6}-\dfrac{3i}{6}[/latex] [latex]\left(\dfrac{1}{2}+2i\right)\left(\dfrac{1}{3}-3i\right)=\dfrac{37}{6}-\dfrac{1}{2}i[/latex] b) [latex](1+i)(1+i)^3(1+i)^{-1}=(1+i)^{1+3-1}=(1+i)^{3}=1+3i+3i^2+i^3[/latex] [latex]i^2=-1[/latex] e [latex]i^3=-i[/latex] [latex](1+i)(1+i)^3(1+i)^{-1}=1+3i+3\cdot(-1)-i=1+2i-3=-2+2i[/latex] c) [latex]3(7+2i)-[(5+4i)+1]=21+6i-[6+4i]=21+6i-6-4i=15+2i[/latex] 11) a) [latex](2-3i)(-2+i)=-4+2i+6i-3i^2=-4+8i+3=-1+8i[/latex] b) [latex](3+i)(3-i)\left(\dfrac{1}{5}+\dfrac{1}{10}i\right)=2+i[/latex] [latex](3+i)(3-i)=3^2-i^2=9-(-1)=10[/latex] [latex](3+i)(3-i)\left(\dfrac{1}{5}+\dfrac{1}{10}i\right)=10\cdot\left(\dfrac{1}{10}+\dfrac{1}{10}i\right)[/latex] [latex](3+i)(3-i)\left(\dfrac{1}{5}+\dfrac{1}{10}i\right)=\dfrac{10}{10}+\dfrac{10}{10}i[/latex] [latex](3+i)(3-i)\left(\dfrac{1}{5}+\dfrac{1}{10}i\right)=1+i[/latex]

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