Vermenigvuldigen en delen (a+bi)

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Bereken

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

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Bereken

Verbetersleutel

1. \((10-8i) \cdot (-10-9i)= -100-90i +80 i+72i^2 = -100-90i +80 i-72= \color{red}{-100-72}\color{blue}{-90i +80i}=\color{red}{-172}\color{blue}{-10i}\)
2. \((-4+8i) \cdot (-1-6i)= 4+24i -8 i-48i^2 = 4+24i -8 i+48= \color{red}{4+48}\color{blue}{+24i -8i}=\color{red}{52}\color{blue}{+16i}\)
3. \(\frac{-5-9i}{5-5i}= \frac{-5-9i}{5-5i} \cdot \frac{5+5i}{5+5i} = \frac{-25-25i -45 i-45i^2 }{(5)^2-(-5i)^2} = \frac{-25-25i -45 i+45}{25 + 25} = \frac{20-70i }{50} = \frac{2}{5} + \frac{-7}{5}i \)
4. \((-4i) \cdot (7+i)= -28 i-4i^2 = \color{red}{4}\color{blue}{-28i}\)
5. \((-2+5i)\cdot (-3i)= +6 i-15i^2 = \color{red}{15}\color{blue}{+6i}\)
6. \(\frac{6-i}{2-5i}= \frac{6-i}{2-5i} \cdot \frac{2+5i}{2+5i} = \frac{12+30i -2 i-5i^2 }{(2)^2-(-5i)^2} = \frac{12+30i -2 i+5}{4 + 25} = \frac{17+28i }{29} = \frac{17}{29} - \frac{-28}{29}i \)
7. \((+i) \cdot (-4+4i)= -4 i+4i^2 = \color{red}{-4}\color{blue}{-4i}\)
8. \(\frac{-3-8i}{-9-9i}= \frac{-3-8i}{-9-9i} \cdot \frac{-9+9i}{-9+9i} = \frac{27-27i +72 i-72i^2 }{(-9)^2-(-9i)^2} = \frac{27-27i +72 i+72}{81 + 81} = \frac{99+45i }{162} = \frac{11}{18} - \frac{-5}{18}i \)
9. \((3-5i) \cdot (7-7i)= 21-21i -35 i+35i^2 = 21-21i -35 i-35= \color{red}{21-35}\color{blue}{-21i -35i}=\color{red}{-14}\color{blue}{-56i}\)
10. \((-8-8i) \cdot (1+2i)= -8-16i -8 i-16i^2 = -8-16i -8 i+16= \color{red}{-8+16}\color{blue}{-16i -8i}=\color{red}{8}\color{blue}{-24i}\)
11. \(\frac{8-8i}{1+8i}= \frac{8-8i}{1+8i} \cdot \frac{1-8i}{1-8i} = \frac{8-64i -8 i+64i^2 }{(1)^2-(8i)^2} = \frac{8-64i -8 i-64}{1 + 64} = \frac{-56-72i }{65} = \frac{-56}{65} + \frac{-72}{65}i \)
12. \((+i) \cdot (1-5i)= +1 i-5i^2 = \color{red}{5}\color{blue}{+i}\)