Vermenigvuldigen en delen (a+bi)

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Bereken

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

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Bereken

Verbetersleutel

1. \(\frac{9-6i}{-4-10i}= \frac{9-6i}{-4-10i} \cdot \frac{-4+10i}{-4+10i} = \frac{-36+90i +24 i-60i^2 }{(-4)^2-(-10i)^2} = \frac{-36+90i +24 i+60}{16 + 100} = \frac{24+114i }{116} = \frac{6}{29} - \frac{-57}{58}i \)
2. \(\frac{-2+i}{3-9i}= \frac{-2+i}{3-9i} \cdot \frac{3+9i}{3+9i} = \frac{-6-18i +3 i+9i^2 }{(3)^2-(-9i)^2} = \frac{-6-18i +3 i-9}{9 + 81} = \frac{-15-15i }{90} = \frac{-1}{6} + \frac{-1}{6}i \)
3. \((-3i) \cdot (-2-9i)= +6 i+27i^2 = \color{red}{-27}\color{blue}{+6i}\)
4. \((6+3i) \cdot (2+7i)= 12+42i +6 i+21i^2 = 12+42i +6 i-21= \color{red}{12-21}\color{blue}{+42i +6i}=\color{red}{-9}\color{blue}{+48i}\)
5. \(\frac{-4-2i}{-4+9i}= \frac{-4-2i}{-4+9i} \cdot \frac{-4-9i}{-4-9i} = \frac{16+36i +8 i+18i^2 }{(-4)^2-(9i)^2} = \frac{16+36i +8 i-18}{16 + 81} = \frac{-2+44i }{97} = \frac{-2}{97} - \frac{-44}{97}i \)
6. \((+i) \cdot (2+10i)= +2 i+10i^2 = \color{red}{-10}\color{blue}{+2i}\)
7. \((-8i) \cdot (-3-6i)= +24 i+48i^2 = \color{red}{-48}\color{blue}{+24i}\)
8. \(\frac{-1-10i}{-10-2i}= \frac{-1-10i}{-10-2i} \cdot \frac{-10+2i}{-10+2i} = \frac{10-2i +100 i-20i^2 }{(-10)^2-(-2i)^2} = \frac{10-2i +100 i+20}{100 + 4} = \frac{30+98i }{104} = \frac{15}{52} - \frac{-49}{52}i \)
9. \((2+10i) \cdot (-10-7i)= -20-14i -100 i-70i^2 = -20-14i -100 i+70= \color{red}{-20+70}\color{blue}{-14i -100i}=\color{red}{50}\color{blue}{-114i}\)
10. \((+9i) \cdot (-8+5i)= -72 i+45i^2 = \color{red}{-45}\color{blue}{-72i}\)
11. \(\frac{-7+10i}{-5+10i}= \frac{-7+10i}{-5+10i} \cdot \frac{-5-10i}{-5-10i} = \frac{35+70i -50 i-100i^2 }{(-5)^2-(10i)^2} = \frac{35+70i -50 i+100}{25 + 100} = \frac{135+20i }{125} = \frac{27}{25} - \frac{-4}{25}i \)
12. \(\frac{2-5i}{-5+2i}= \frac{2-5i}{-5+2i} \cdot \frac{-5-2i}{-5-2i} = \frac{-10-4i +25 i+10i^2 }{(-5)^2-(2i)^2} = \frac{-10-4i +25 i-10}{25 + 4} = \frac{-20+21i }{29} = \frac{-20}{29} - \frac{-21}{29}i \)