Vermenigvuldigen en delen (a+bi)

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Bereken

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

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Bereken

Verbetersleutel

1. \((1-5i) \cdot (-5+10i)= -5+10i +25 i-50i^2 = -5+10i +25 i+50= \color{red}{-5+50}\color{blue}{+10i +25i}=\color{red}{45}\color{blue}{+35i}\)
2. \((1-4i) \cdot (5+3i)= 5+3i -20 i-12i^2 = 5+3i -20 i+12= \color{red}{5+12}\color{blue}{+3i -20i}=\color{red}{17}\color{blue}{-17i}\)
3. \((2-i)\cdot (-5i)= -10 i+5i^2 = \color{red}{-5}\color{blue}{-10i}\)
4. \((5+3i) \cdot (-3+8i)= -15+40i -9 i+24i^2 = -15+40i -9 i-24= \color{red}{-15-24}\color{blue}{+40i -9i}=\color{red}{-39}\color{blue}{+31i}\)
5. \(\frac{5-4i}{-10+3i}= \frac{5-4i}{-10+3i} \cdot \frac{-10-3i}{-10-3i} = \frac{-50-15i +40 i+12i^2 }{(-10)^2-(3i)^2} = \frac{-50-15i +40 i-12}{100 + 9} = \frac{-62+25i }{109} = \frac{-62}{109} - \frac{-25}{109}i \)
6. \(\frac{2+10i}{-3-8i}= \frac{2+10i}{-3-8i} \cdot \frac{-3+8i}{-3+8i} = \frac{-6+16i -30 i+80i^2 }{(-3)^2-(-8i)^2} = \frac{-6+16i -30 i-80}{9 + 64} = \frac{-86-14i }{73} = \frac{-86}{73} + \frac{-14}{73}i \)
7. \(\frac{3-8i}{-2+6i}= \frac{3-8i}{-2+6i} \cdot \frac{-2-6i}{-2-6i} = \frac{-6-18i +16 i+48i^2 }{(-2)^2-(6i)^2} = \frac{-6-18i +16 i-48}{4 + 36} = \frac{-54-2i }{40} = \frac{-27}{20} + \frac{-1}{20}i \)
8. \(\frac{5+9i}{6-10i}= \frac{5+9i}{6-10i} \cdot \frac{6+10i}{6+10i} = \frac{30+50i +54 i+90i^2 }{(6)^2-(-10i)^2} = \frac{30+50i +54 i-90}{36 + 100} = \frac{-60+104i }{136} = \frac{-15}{34} - \frac{-13}{17}i \)
9. \((-6+2i) \cdot (-6+9i)= 36-54i -12 i+18i^2 = 36-54i -12 i-18= \color{red}{36-18}\color{blue}{-54i -12i}=\color{red}{18}\color{blue}{-66i}\)
10. \(\frac{4+7i}{-4-10i}= \frac{4+7i}{-4-10i} \cdot \frac{-4+10i}{-4+10i} = \frac{-16+40i -28 i+70i^2 }{(-4)^2-(-10i)^2} = \frac{-16+40i -28 i-70}{16 + 100} = \frac{-86+12i }{116} = \frac{-43}{58} - \frac{-3}{29}i \)
11. \((4-4i) \cdot (-3+10i)= -12+40i +12 i-40i^2 = -12+40i +12 i+40= \color{red}{-12+40}\color{blue}{+40i +12i}=\color{red}{28}\color{blue}{+52i}\)
12. \(\frac{5+3i}{9-6i}= \frac{5+3i}{9-6i} \cdot \frac{9+6i}{9+6i} = \frac{45+30i +27 i+18i^2 }{(9)^2-(-6i)^2} = \frac{45+30i +27 i-18}{81 + 36} = \frac{27+57i }{117} = \frac{3}{13} - \frac{-19}{39}i \)