Vermenigvuldigen en delen (a+bi)

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Bereken

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

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Bereken

Verbetersleutel

1. \(\frac{10+2i}{-5-8i}= \frac{10+2i}{-5-8i} \cdot \frac{-5+8i}{-5+8i} = \frac{-50+80i -10 i+16i^2 }{(-5)^2-(-8i)^2} = \frac{-50+80i -10 i-16}{25 + 64} = \frac{-66+70i }{89} = \frac{-66}{89} - \frac{-70}{89}i \)
2. \((-7+10i) \cdot (3+i)= -21-7i +30 i+10i^2 = -21-7i +30 i-10= \color{red}{-21-10}\color{blue}{-7i +30i}=\color{red}{-31}\color{blue}{+23i}\)
3. \(\frac{-5-10i}{-4-i}= \frac{-5-10i}{-4-i} \cdot \frac{-4+i}{-4+i} = \frac{20-5i +40 i-10i^2 }{(-4)^2-(-1i)^2} = \frac{20-5i +40 i+10}{16 + 1} = \frac{30+35i }{17} = \frac{30}{17} - \frac{-35}{17}i \)
4. \((-4-4i) \cdot (-1-5i)= 4+20i +4 i+20i^2 = 4+20i +4 i-20= \color{red}{4-20}\color{blue}{+20i +4i}=\color{red}{-16}\color{blue}{+24i}\)
5. \((-10+3i) \cdot (-6+6i)= 60-60i -18 i+18i^2 = 60-60i -18 i-18= \color{red}{60-18}\color{blue}{-60i -18i}=\color{red}{42}\color{blue}{-78i}\)
6. \((-9-5i)\cdot (-6i)= +54 i+30i^2 = \color{red}{-30}\color{blue}{+54i}\)
7. \((-1+5i)\cdot (-2i)= +2 i-10i^2 = \color{red}{10}\color{blue}{+2i}\)
8. \((-9+6i)\cdot (+4i)= -36 i+24i^2 = \color{red}{-24}\color{blue}{-36i}\)
9. \((-10+3i)\cdot (+7i)= -70 i+21i^2 = \color{red}{-21}\color{blue}{-70i}\)
10. \(\frac{3+i}{4-8i}= \frac{3+i}{4-8i} \cdot \frac{4+8i}{4+8i} = \frac{12+24i +4 i+8i^2 }{(4)^2-(-8i)^2} = \frac{12+24i +4 i-8}{16 + 64} = \frac{4+28i }{80} = \frac{1}{20} - \frac{-7}{20}i \)
11. \(\frac{3+2i}{5+3i}= \frac{3+2i}{5+3i} \cdot \frac{5-3i}{5-3i} = \frac{15-9i +10 i-6i^2 }{(5)^2-(3i)^2} = \frac{15-9i +10 i+6}{25 + 9} = \frac{21+i }{34} = \frac{21}{34} - \frac{-1}{34}i \)
12. \((7-i) \cdot (-10-i)= -70-7i +10 i+i^2 = -70-7i +10 i-= \color{red}{-70-1}\color{blue}{-7i +10i}=\color{red}{-71}\color{blue}{+3i}\)