Basisbewerkingen gemengd (a+bi)

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Bereken

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

Bereken

Verbetersleutel

  1. \((-i) \cdot (6+2i)= -6 i-2i^2 = \color{red}{2}\color{blue}{-6i}\)
  2. \((10-i)+(-3-2i)= 10-i -3-2i =\color{red}{10-3}\color{blue}{-i -2i}=\color{red}{7}\color{blue}{-3i}\)
  3. \((-4-5i)-(-5-5i)= -4-5i +5+5i =\color{red}{-4+5}\color{blue}{-5i +5i}=\color{red}{1}\)
  4. \(\frac{-10+5i}{1-i}= \frac{-10+5i}{1-i} \cdot \frac{1+i}{1+i} = \frac{-10-10i +5 i+5i^2 }{(1)^2-(-1i)^2} = \frac{-10-10i +5 i-5}{1 + 1} = \frac{-15-5i }{2} = \frac{-15}{2} + \frac{-5}{2}i \)
  5. \((-i) \cdot (-3+3i)= +3 i-3i^2 = \color{red}{3}\color{blue}{+3i}\)
  6. \((3+i) \cdot (-5+2i)= -15+6i -5 i+2i^2 = -15+6i -5 i-2= \color{red}{-15-2}\color{blue}{+6i -5i}=\color{red}{-17}\color{blue}{+i}\)
  7. \(\frac{10-10i}{-5-10i}= \frac{10-10i}{-5-10i} \cdot \frac{-5+10i}{-5+10i} = \frac{-50+100i +50 i-100i^2 }{(-5)^2-(-10i)^2} = \frac{-50+100i +50 i+100}{25 + 100} = \frac{50+150i }{125} = \frac{2}{5} - \frac{-6}{5}i \)
  8. \((3-6i)+(2-7i)= 3-6i +2-7i =\color{red}{3+2}\color{blue}{-6i -7i}=\color{red}{5}\color{blue}{-13i}\)
  9. \(\frac{-5-5i}{6+4i}= \frac{-5-5i}{6+4i} \cdot \frac{6-4i}{6-4i} = \frac{-30+20i -30 i+20i^2 }{(6)^2-(4i)^2} = \frac{-30+20i -30 i-20}{36 + 16} = \frac{-50-10i }{52} = \frac{-25}{26} + \frac{-5}{26}i \)
  10. \((1+i)\cdot (-7i)= -7 i-7i^2 = \color{red}{7}\color{blue}{-7i}\)
  11. \((7-3i)\cdot (+i)= +7 i-3i^2 = \color{red}{3}\color{blue}{+7i}\)
  12. \((-7+4i)+(3+6i)= -7+4i +3+6i =\color{red}{-7+3}\color{blue}{+4i +6i}=\color{red}{-4}\color{blue}{+10i}\)
Oefeningengenerator wiskundeoefeningen.be 2026-05-08 00:26:51
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