Vermenigvuldigen en delen (a+bi)

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Bereken

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

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Bereken

Verbetersleutel

1. \(\frac{10+9i}{3+2i}= \frac{10+9i}{3+2i} \cdot \frac{3-2i}{3-2i} = \frac{30-20i +27 i-18i^2 }{(3)^2-(2i)^2} = \frac{30-20i +27 i+18}{9 + 4} = \frac{48+7i }{13} = \frac{48}{13} - \frac{-7}{13}i \)
2. \((-8i) \cdot (-1-3i)= +8 i+24i^2 = \color{red}{-24}\color{blue}{+8i}\)
3. \(\frac{7+8i}{1-3i}= \frac{7+8i}{1-3i} \cdot \frac{1+3i}{1+3i} = \frac{7+21i +8 i+24i^2 }{(1)^2-(-3i)^2} = \frac{7+21i +8 i-24}{1 + 9} = \frac{-17+29i }{10} = \frac{-17}{10} - \frac{-29}{10}i \)
4. \((+3i) \cdot (-5+6i)= -15 i+18i^2 = \color{red}{-18}\color{blue}{-15i}\)
5. \((1+3i)\cdot (+6i)= +6 i+18i^2 = \color{red}{-18}\color{blue}{+6i}\)
6. \((7+7i) \cdot (-7-10i)= -49-70i -49 i-70i^2 = -49-70i -49 i+70= \color{red}{-49+70}\color{blue}{-70i -49i}=\color{red}{21}\color{blue}{-119i}\)
7. \((-9-5i) \cdot (10+i)= -90-9i -50 i-5i^2 = -90-9i -50 i+5= \color{red}{-90+5}\color{blue}{-9i -50i}=\color{red}{-85}\color{blue}{-59i}\)
8. \(\frac{6-7i}{-6-3i}= \frac{6-7i}{-6-3i} \cdot \frac{-6+3i}{-6+3i} = \frac{-36+18i +42 i-21i^2 }{(-6)^2-(-3i)^2} = \frac{-36+18i +42 i+21}{36 + 9} = \frac{-15+60i }{45} = \frac{-1}{3} - \frac{-4}{3}i \)
9. \(\frac{7+2i}{-5-i}= \frac{7+2i}{-5-i} \cdot \frac{-5+i}{-5+i} = \frac{-35+7i -10 i+2i^2 }{(-5)^2-(-1i)^2} = \frac{-35+7i -10 i-2}{25 + 1} = \frac{-37-3i }{26} = \frac{-37}{26} + \frac{-3}{26}i \)
10. \(\frac{4-2i}{9-6i}= \frac{4-2i}{9-6i} \cdot \frac{9+6i}{9+6i} = \frac{36+24i -18 i-12i^2 }{(9)^2-(-6i)^2} = \frac{36+24i -18 i+12}{81 + 36} = \frac{48+6i }{117} = \frac{16}{39} - \frac{-2}{39}i \)
11. \((-8-5i) \cdot (-3-9i)= 24+72i +15 i+45i^2 = 24+72i +15 i-45= \color{red}{24-45}\color{blue}{+72i +15i}=\color{red}{-21}\color{blue}{+87i}\)
12. \(\frac{-4-8i}{-3+10i}= \frac{-4-8i}{-3+10i} \cdot \frac{-3-10i}{-3-10i} = \frac{12+40i +24 i+80i^2 }{(-3)^2-(10i)^2} = \frac{12+40i +24 i-80}{9 + 100} = \frac{-68+64i }{109} = \frac{-68}{109} - \frac{-64}{109}i \)