Vermenigvuldigen en delen (a+bi)

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Bereken

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

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Bereken

Verbetersleutel

1. \((+3i) \cdot (9-2i)= +27 i-6i^2 = \color{red}{6}\color{blue}{+27i}\)
2. \((5-8i) \cdot (-3-8i)= -15-40i +24 i+64i^2 = -15-40i +24 i-64= \color{red}{-15-64}\color{blue}{-40i +24i}=\color{red}{-79}\color{blue}{-16i}\)
3. \((-7-5i) \cdot (-4-6i)= 28+42i +20 i+30i^2 = 28+42i +20 i-30= \color{red}{28-30}\color{blue}{+42i +20i}=\color{red}{-2}\color{blue}{+62i}\)
4. \((7+7i)\cdot (-9i)= -63 i-63i^2 = \color{red}{63}\color{blue}{-63i}\)
5. \((4+4i) \cdot (1-2i)= 4-8i +4 i-8i^2 = 4-8i +4 i+8= \color{red}{4+8}\color{blue}{-8i +4i}=\color{red}{12}\color{blue}{-4i}\)
6. \((6+2i) \cdot (4-10i)= 24-60i +8 i-20i^2 = 24-60i +8 i+20= \color{red}{24+20}\color{blue}{-60i +8i}=\color{red}{44}\color{blue}{-52i}\)
7. \(\frac{-10+4i}{-2-2i}= \frac{-10+4i}{-2-2i} \cdot \frac{-2+2i}{-2+2i} = \frac{20-20i -8 i+8i^2 }{(-2)^2-(-2i)^2} = \frac{20-20i -8 i-8}{4 + 4} = \frac{12-28i }{8} = \frac{3}{2} + \frac{-7}{2}i \)
8. \(\frac{-10+2i}{-8+6i}= \frac{-10+2i}{-8+6i} \cdot \frac{-8-6i}{-8-6i} = \frac{80+60i -16 i-12i^2 }{(-8)^2-(6i)^2} = \frac{80+60i -16 i+12}{64 + 36} = \frac{92+44i }{100} = \frac{23}{25} - \frac{-11}{25}i \)
9. \(\frac{2-7i}{-3+10i}= \frac{2-7i}{-3+10i} \cdot \frac{-3-10i}{-3-10i} = \frac{-6-20i +21 i+70i^2 }{(-3)^2-(10i)^2} = \frac{-6-20i +21 i-70}{9 + 100} = \frac{-76+i }{109} = \frac{-76}{109} - \frac{-1}{109}i \)
10. \((-10+10i) \cdot (-6-4i)= 60+40i -60 i-40i^2 = 60+40i -60 i+40= \color{red}{60+40}\color{blue}{+40i -60i}=\color{red}{100}\color{blue}{-20i}\)
11. \(\frac{10+9i}{-4-5i}= \frac{10+9i}{-4-5i} \cdot \frac{-4+5i}{-4+5i} = \frac{-40+50i -36 i+45i^2 }{(-4)^2-(-5i)^2} = \frac{-40+50i -36 i-45}{16 + 25} = \frac{-85+14i }{41} = \frac{-85}{41} - \frac{-14}{41}i \)
12. \(\frac{5+5i}{8+i}= \frac{5+5i}{8+i} \cdot \frac{8-i}{8-i} = \frac{40-5i +40 i-5i^2 }{(8)^2-(1i)^2} = \frac{40-5i +40 i+5}{64 + 1} = \frac{45+35i }{65} = \frac{9}{13} - \frac{-7}{13}i \)