Vermenigvuldigen en delen (a+bi)

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Bereken

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

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Bereken

Verbetersleutel

1. \(\frac{5-2i}{4+7i}= \frac{5-2i}{4+7i} \cdot \frac{4-7i}{4-7i} = \frac{20-35i -8 i+14i^2 }{(4)^2-(7i)^2} = \frac{20-35i -8 i-14}{16 + 49} = \frac{6-43i }{65} = \frac{6}{65} + \frac{-43}{65}i \)
2. \((-8i) \cdot (-7+3i)= +56 i-24i^2 = \color{red}{24}\color{blue}{+56i}\)
3. \(\frac{-10+6i}{-4+9i}= \frac{-10+6i}{-4+9i} \cdot \frac{-4-9i}{-4-9i} = \frac{40+90i -24 i-54i^2 }{(-4)^2-(9i)^2} = \frac{40+90i -24 i+54}{16 + 81} = \frac{94+66i }{97} = \frac{94}{97} - \frac{-66}{97}i \)
4. \((-2i) \cdot (6+7i)= -12 i-14i^2 = \color{red}{14}\color{blue}{-12i}\)
5. \(\frac{1+6i}{3+6i}= \frac{1+6i}{3+6i} \cdot \frac{3-6i}{3-6i} = \frac{3-6i +18 i-36i^2 }{(3)^2-(6i)^2} = \frac{3-6i +18 i+36}{9 + 36} = \frac{39+12i }{45} = \frac{13}{15} - \frac{-4}{15}i \)
6. \((-5i) \cdot (4-9i)= -20 i+45i^2 = \color{red}{-45}\color{blue}{-20i}\)
7. \(\frac{3+2i}{-10+2i}= \frac{3+2i}{-10+2i} \cdot \frac{-10-2i}{-10-2i} = \frac{-30-6i -20 i-4i^2 }{(-10)^2-(2i)^2} = \frac{-30-6i -20 i+4}{100 + 4} = \frac{-26-26i }{104} = \frac{-1}{4} + \frac{-1}{4}i \)
8. \(\frac{5-i}{4-3i}= \frac{5-i}{4-3i} \cdot \frac{4+3i}{4+3i} = \frac{20+15i -4 i-3i^2 }{(4)^2-(-3i)^2} = \frac{20+15i -4 i+3}{16 + 9} = \frac{23+11i }{25} = \frac{23}{25} - \frac{-11}{25}i \)
9. \((6-5i)\cdot (-7i)= -42 i+35i^2 = \color{red}{-35}\color{blue}{-42i}\)
10. \((5+3i) \cdot (-4+2i)= -20+10i -12 i+6i^2 = -20+10i -12 i-6= \color{red}{-20-6}\color{blue}{+10i -12i}=\color{red}{-26}\color{blue}{-2i}\)
11. \((+6i) \cdot (-4-9i)= -24 i-54i^2 = \color{red}{54}\color{blue}{-24i}\)
12. \((-9-8i) \cdot (3-3i)= -27+27i -24 i+24i^2 = -27+27i -24 i-24= \color{red}{-27-24}\color{blue}{+27i -24i}=\color{red}{-51}\color{blue}{+3i}\)