ayresrafael10
3

# a pergunta e resposta da 10 e11

(1) Respostas
Jailson123

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