Vermenigvuldigen en delen (a+bi)

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Bereken

1. \((-1+8i) \cdot (-10+4i)\)
2. \((6+5i) \cdot (-10+5i)\)
3. \(\frac{-7+4i}{5+6i}\)
4. \(\frac{2+9i}{-3+9i}\)
5. \(\frac{-9-4i}{-7-9i}\)
6. \(\frac{-10+9i}{-7+3i}\)
7. \(\frac{10+i}{4+10i}\)
8. \((6-6i) \cdot (-5-6i)\)
9. \(\frac{-1-10i}{10+2i}\)
10. \((-4+10i)\cdot (-4i)\)
11. \(\frac{7-6i}{-9+9i}\)
12. \((-8+10i)\cdot (-5i)\)

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Bereken

Verbetersleutel

1. \((-1+8i) \cdot (-10+4i)= 10-4i -80 i+32i^2 = 10-4i -80 i-32= \color{red}{10-32}\color{blue}{-4i -80i}=\color{red}{-22}\color{blue}{-84i}\)
2. \((6+5i) \cdot (-10+5i)= -60+30i -50 i+25i^2 = -60+30i -50 i-25= \color{red}{-60-25}\color{blue}{+30i -50i}=\color{red}{-85}\color{blue}{-20i}\)
3. \(\frac{-7+4i}{5+6i}= \frac{-7+4i}{5+6i} \cdot \frac{5-6i}{5-6i} = \frac{-35+42i +20 i-24i^2 }{(5)^2-(6i)^2} = \frac{-35+42i +20 i+24}{25 + 36} = \frac{-11+62i }{61} = \frac{-11}{61} - \frac{-62}{61}i \)
4. \(\frac{2+9i}{-3+9i}= \frac{2+9i}{-3+9i} \cdot \frac{-3-9i}{-3-9i} = \frac{-6-18i -27 i-81i^2 }{(-3)^2-(9i)^2} = \frac{-6-18i -27 i+81}{9 + 81} = \frac{75-45i }{90} = \frac{5}{6} + \frac{-1}{2}i \)
5. \(\frac{-9-4i}{-7-9i}= \frac{-9-4i}{-7-9i} \cdot \frac{-7+9i}{-7+9i} = \frac{63-81i +28 i-36i^2 }{(-7)^2-(-9i)^2} = \frac{63-81i +28 i+36}{49 + 81} = \frac{99-53i }{130} = \frac{99}{130} + \frac{-53}{130}i \)
6. \(\frac{-10+9i}{-7+3i}= \frac{-10+9i}{-7+3i} \cdot \frac{-7-3i}{-7-3i} = \frac{70+30i -63 i-27i^2 }{(-7)^2-(3i)^2} = \frac{70+30i -63 i+27}{49 + 9} = \frac{97-33i }{58} = \frac{97}{58} + \frac{-33}{58}i \)
7. \(\frac{10+i}{4+10i}= \frac{10+i}{4+10i} \cdot \frac{4-10i}{4-10i} = \frac{40-100i +4 i-10i^2 }{(4)^2-(10i)^2} = \frac{40-100i +4 i+10}{16 + 100} = \frac{50-96i }{116} = \frac{25}{58} + \frac{-24}{29}i \)
8. \((6-6i) \cdot (-5-6i)= -30-36i +30 i+36i^2 = -30-36i +30 i-36= \color{red}{-30-36}\color{blue}{-36i +30i}=\color{red}{-66}\color{blue}{-6i}\)
9. \(\frac{-1-10i}{10+2i}= \frac{-1-10i}{10+2i} \cdot \frac{10-2i}{10-2i} = \frac{-10+2i -100 i+20i^2 }{(10)^2-(2i)^2} = \frac{-10+2i -100 i-20}{100 + 4} = \frac{-30-98i }{104} = \frac{-15}{52} + \frac{-49}{52}i \)
10. \((-4+10i)\cdot (-4i)= +16 i-40i^2 = \color{red}{40}\color{blue}{+16i}\)
11. \(\frac{7-6i}{-9+9i}= \frac{7-6i}{-9+9i} \cdot \frac{-9-9i}{-9-9i} = \frac{-63-63i +54 i+54i^2 }{(-9)^2-(9i)^2} = \frac{-63-63i +54 i-54}{81 + 81} = \frac{-117-9i }{162} = \frac{-13}{18} + \frac{-1}{18}i \)
12. \((-8+10i)\cdot (-5i)= +40 i-50i^2 = \color{red}{50}\color{blue}{+40i}\)