Vermenigvuldigen en delen (a+bi)

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Bereken

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

Bereken

Verbetersleutel

  1. \((-9+10i)\cdot (-10i)= +90 i-100i^2 = \color{red}{100}\color{blue}{+90i}\)
  2. \((-6-4i) \cdot (4+6i)= -24-36i -16 i-24i^2 = -24-36i -16 i+24= \color{red}{-24+24}\color{blue}{-36i -16i}=\color{blue}{-52i}\)
  3. \(\frac{-2-5i}{9-8i}= \frac{-2-5i}{9-8i} \cdot \frac{9+8i}{9+8i} = \frac{-18-16i -45 i-40i^2 }{(9)^2-(-8i)^2} = \frac{-18-16i -45 i+40}{81 + 64} = \frac{22-61i }{145} = \frac{22}{145} + \frac{-61}{145}i \)
  4. \(\frac{-6+10i}{1+4i}= \frac{-6+10i}{1+4i} \cdot \frac{1-4i}{1-4i} = \frac{-6+24i +10 i-40i^2 }{(1)^2-(4i)^2} = \frac{-6+24i +10 i+40}{1 + 16} = \frac{34+34i }{17} = 2- -2i\)
  5. \((1+8i)\cdot (-3i)= -3 i-24i^2 = \color{red}{24}\color{blue}{-3i}\)
  6. \((3+8i) \cdot (7-7i)= 21-21i +56 i-56i^2 = 21-21i +56 i+56= \color{red}{21+56}\color{blue}{-21i +56i}=\color{red}{77}\color{blue}{+35i}\)
  7. \(\frac{-7-3i}{-9-6i}= \frac{-7-3i}{-9-6i} \cdot \frac{-9+6i}{-9+6i} = \frac{63-42i +27 i-18i^2 }{(-9)^2-(-6i)^2} = \frac{63-42i +27 i+18}{81 + 36} = \frac{81-15i }{117} = \frac{9}{13} + \frac{-5}{39}i \)
  8. \((-8i) \cdot (-7+3i)= +56 i-24i^2 = \color{red}{24}\color{blue}{+56i}\)
  9. \(\frac{-6-4i}{-7-4i}= \frac{-6-4i}{-7-4i} \cdot \frac{-7+4i}{-7+4i} = \frac{42-24i +28 i-16i^2 }{(-7)^2-(-4i)^2} = \frac{42-24i +28 i+16}{49 + 16} = \frac{58+4i }{65} = \frac{58}{65} - \frac{-4}{65}i \)
  10. \((-5i) \cdot (-4-4i)= +20 i+20i^2 = \color{red}{-20}\color{blue}{+20i}\)
  11. \((5+i) \cdot (8-5i)= 40-25i +8 i-5i^2 = 40-25i +8 i+5= \color{red}{40+5}\color{blue}{-25i +8i}=\color{red}{45}\color{blue}{-17i}\)
  12. \(\frac{2-8i}{4+8i}= \frac{2-8i}{4+8i} \cdot \frac{4-8i}{4-8i} = \frac{8-16i -32 i+64i^2 }{(4)^2-(8i)^2} = \frac{8-16i -32 i-64}{16 + 64} = \frac{-56-48i }{80} = \frac{-7}{10} + \frac{-3}{5}i \)
Oefeningengenerator wiskundeoefeningen.be 2025-07-09 10:35:13
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