Vermenigvuldigen en delen (a+bi)

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Bereken

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

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Bereken

Verbetersleutel

1. \((7-9i) \cdot (-5+9i)= -35+63i +45 i-81i^2 = -35+63i +45 i+81= \color{red}{-35+81}\color{blue}{+63i +45i}=\color{red}{46}\color{blue}{+108i}\)
2. \(\frac{10-i}{-2-7i}= \frac{10-i}{-2-7i} \cdot \frac{-2+7i}{-2+7i} = \frac{-20+70i +2 i-7i^2 }{(-2)^2-(-7i)^2} = \frac{-20+70i +2 i+7}{4 + 49} = \frac{-13+72i }{53} = \frac{-13}{53} - \frac{-72}{53}i \)
3. \(\frac{1-10i}{10-3i}= \frac{1-10i}{10-3i} \cdot \frac{10+3i}{10+3i} = \frac{10+3i -100 i-30i^2 }{(10)^2-(-3i)^2} = \frac{10+3i -100 i+30}{100 + 9} = \frac{40-97i }{109} = \frac{40}{109} + \frac{-97}{109}i \)
4. \((2+9i) \cdot (-8-7i)= -16-14i -72 i-63i^2 = -16-14i -72 i+63= \color{red}{-16+63}\color{blue}{-14i -72i}=\color{red}{47}\color{blue}{-86i}\)
5. \((6+7i)\cdot (-2i)= -12 i-14i^2 = \color{red}{14}\color{blue}{-12i}\)
6. \((-1-3i) \cdot (5+3i)= -5-3i -15 i-9i^2 = -5-3i -15 i+9= \color{red}{-5+9}\color{blue}{-3i -15i}=\color{red}{4}\color{blue}{-18i}\)
7. \(\frac{-1-8i}{-4+9i}= \frac{-1-8i}{-4+9i} \cdot \frac{-4-9i}{-4-9i} = \frac{4+9i +32 i+72i^2 }{(-4)^2-(9i)^2} = \frac{4+9i +32 i-72}{16 + 81} = \frac{-68+41i }{97} = \frac{-68}{97} - \frac{-41}{97}i \)
8. \((9-7i)\cdot (+8i)= +72 i-56i^2 = \color{red}{56}\color{blue}{+72i}\)
9. \((-7-9i) \cdot (-7-8i)= 49+56i +63 i+72i^2 = 49+56i +63 i-72= \color{red}{49-72}\color{blue}{+56i +63i}=\color{red}{-23}\color{blue}{+119i}\)
10. \(\frac{4+5i}{3-4i}= \frac{4+5i}{3-4i} \cdot \frac{3+4i}{3+4i} = \frac{12+16i +15 i+20i^2 }{(3)^2-(-4i)^2} = \frac{12+16i +15 i-20}{9 + 16} = \frac{-8+31i }{25} = \frac{-8}{25} - \frac{-31}{25}i \)
11. \((-1-9i) \cdot (1-2i)= -1+2i -9 i+18i^2 = -1+2i -9 i-18= \color{red}{-1-18}\color{blue}{+2i -9i}=\color{red}{-19}\color{blue}{-7i}\)
12. \(\frac{-1+9i}{1+6i}= \frac{-1+9i}{1+6i} \cdot \frac{1-6i}{1-6i} = \frac{-1+6i +9 i-54i^2 }{(1)^2-(6i)^2} = \frac{-1+6i +9 i+54}{1 + 36} = \frac{53+15i }{37} = \frac{53}{37} - \frac{-15}{37}i \)