Vermenigvuldigen en delen (a+bi)

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Bereken

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

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Bereken

Verbetersleutel

1. \((-10i) \cdot (-7-2i)= +70 i+20i^2 = \color{red}{-20}\color{blue}{+70i}\)
2. \(\frac{3-2i}{1-4i}= \frac{3-2i}{1-4i} \cdot \frac{1+4i}{1+4i} = \frac{3+12i -2 i-8i^2 }{(1)^2-(-4i)^2} = \frac{3+12i -2 i+8}{1 + 16} = \frac{11+10i }{17} = \frac{11}{17} - \frac{-10}{17}i \)
3. \((+2i) \cdot (3+i)= +6 i+2i^2 = \color{red}{-2}\color{blue}{+6i}\)
4. \((-10i) \cdot (9+i)= -90 i-10i^2 = \color{red}{10}\color{blue}{-90i}\)
5. \(\frac{-1+5i}{-3-4i}= \frac{-1+5i}{-3-4i} \cdot \frac{-3+4i}{-3+4i} = \frac{3-4i -15 i+20i^2 }{(-3)^2-(-4i)^2} = \frac{3-4i -15 i-20}{9 + 16} = \frac{-17-19i }{25} = \frac{-17}{25} + \frac{-19}{25}i \)
6. \(\frac{2-10i}{-2+i}= \frac{2-10i}{-2+i} \cdot \frac{-2-i}{-2-i} = \frac{-4-2i +20 i+10i^2 }{(-2)^2-(1i)^2} = \frac{-4-2i +20 i-10}{4 + 1} = \frac{-14+18i }{5} = \frac{-14}{5} - \frac{-18}{5}i \)
7. \((-7-2i) \cdot (6-7i)= -42+49i -12 i+14i^2 = -42+49i -12 i-14= \color{red}{-42-14}\color{blue}{+49i -12i}=\color{red}{-56}\color{blue}{+37i}\)
8. \(\frac{-1+7i}{9+2i}= \frac{-1+7i}{9+2i} \cdot \frac{9-2i}{9-2i} = \frac{-9+2i +63 i-14i^2 }{(9)^2-(2i)^2} = \frac{-9+2i +63 i+14}{81 + 4} = \frac{5+65i }{85} = \frac{1}{17} - \frac{-13}{17}i \)
9. \(\frac{-8-5i}{-10-10i}= \frac{-8-5i}{-10-10i} \cdot \frac{-10+10i}{-10+10i} = \frac{80-80i +50 i-50i^2 }{(-10)^2-(-10i)^2} = \frac{80-80i +50 i+50}{100 + 100} = \frac{130-30i }{200} = \frac{13}{20} + \frac{-3}{20}i \)
10. \(\frac{-7+i}{10-4i}= \frac{-7+i}{10-4i} \cdot \frac{10+4i}{10+4i} = \frac{-70-28i +10 i+4i^2 }{(10)^2-(-4i)^2} = \frac{-70-28i +10 i-4}{100 + 16} = \frac{-74-18i }{116} = \frac{-37}{58} + \frac{-9}{58}i \)
11. \((6+i) \cdot (-2-5i)= -12-30i -2 i-5i^2 = -12-30i -2 i+5= \color{red}{-12+5}\color{blue}{-30i -2i}=\color{red}{-7}\color{blue}{-32i}\)
12. \(\frac{9+5i}{-9+5i}= \frac{9+5i}{-9+5i} \cdot \frac{-9-5i}{-9-5i} = \frac{-81-45i -45 i-25i^2 }{(-9)^2-(5i)^2} = \frac{-81-45i -45 i+25}{81 + 25} = \frac{-56-90i }{106} = \frac{-28}{53} + \frac{-45}{53}i \)