Vermenigvuldigen en delen (a+bi)

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Bereken

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

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Bereken

Verbetersleutel

1. \((6+7i)\cdot (-i)= -6 i-7i^2 = \color{red}{7}\color{blue}{-6i}\)
2. \(\frac{6-i}{2-6i}= \frac{6-i}{2-6i} \cdot \frac{2+6i}{2+6i} = \frac{12+36i -2 i-6i^2 }{(2)^2-(-6i)^2} = \frac{12+36i -2 i+6}{4 + 36} = \frac{18+34i }{40} = \frac{9}{20} - \frac{-17}{20}i \)
3. \((1+3i) \cdot (-2+10i)= -2+10i -6 i+30i^2 = -2+10i -6 i-30= \color{red}{-2-30}\color{blue}{+10i -6i}=\color{red}{-32}\color{blue}{+4i}\)
4. \(\frac{6+3i}{10+3i}= \frac{6+3i}{10+3i} \cdot \frac{10-3i}{10-3i} = \frac{60-18i +30 i-9i^2 }{(10)^2-(3i)^2} = \frac{60-18i +30 i+9}{100 + 9} = \frac{69+12i }{109} = \frac{69}{109} - \frac{-12}{109}i \)
5. \((7+8i) \cdot (3-4i)= 21-28i +24 i-32i^2 = 21-28i +24 i+32= \color{red}{21+32}\color{blue}{-28i +24i}=\color{red}{53}\color{blue}{-4i}\)
6. \((-2+6i)\cdot (+4i)= -8 i+24i^2 = \color{red}{-24}\color{blue}{-8i}\)
7. \((9-i) \cdot (8+4i)= 72+36i -8 i-4i^2 = 72+36i -8 i+4= \color{red}{72+4}\color{blue}{+36i -8i}=\color{red}{76}\color{blue}{+28i}\)
8. \(\frac{9+9i}{-5-2i}= \frac{9+9i}{-5-2i} \cdot \frac{-5+2i}{-5+2i} = \frac{-45+18i -45 i+18i^2 }{(-5)^2-(-2i)^2} = \frac{-45+18i -45 i-18}{25 + 4} = \frac{-63-27i }{29} = \frac{-63}{29} + \frac{-27}{29}i \)
9. \((+8i) \cdot (4-2i)= +32 i-16i^2 = \color{red}{16}\color{blue}{+32i}\)
10. \(\frac{-6+5i}{-3-6i}= \frac{-6+5i}{-3-6i} \cdot \frac{-3+6i}{-3+6i} = \frac{18-36i -15 i+30i^2 }{(-3)^2-(-6i)^2} = \frac{18-36i -15 i-30}{9 + 36} = \frac{-12-51i }{45} = \frac{-4}{15} + \frac{-17}{15}i \)
11. \((-7i) \cdot (5+3i)= -35 i-21i^2 = \color{red}{21}\color{blue}{-35i}\)
12. \((-1-9i) \cdot (-7-2i)= 7+2i +63 i+18i^2 = 7+2i +63 i-18= \color{red}{7-18}\color{blue}{+2i +63i}=\color{red}{-11}\color{blue}{+65i}\)