Vermenigvuldigen en delen (a+bi)

Hoofdmenu Eentje per keer 

Bereken

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

Bereken

Verbetersleutel

  1. \(\frac{-10+8i}{-7-5i}= \frac{-10+8i}{-7-5i} \cdot \frac{-7+5i}{-7+5i} = \frac{70-50i -56 i+40i^2 }{(-7)^2-(-5i)^2} = \frac{70-50i -56 i-40}{49 + 25} = \frac{30-106i }{74} = \frac{15}{37} + \frac{-53}{37}i \)
  2. \((6+4i)\cdot (-5i)= -30 i-20i^2 = \color{red}{20}\color{blue}{-30i}\)
  3. \((-7-7i)\cdot (+9i)= -63 i-63i^2 = \color{red}{63}\color{blue}{-63i}\)
  4. \((+2i) \cdot (4-10i)= +8 i-20i^2 = \color{red}{20}\color{blue}{+8i}\)
  5. \((2+8i) \cdot (-3+6i)= -6+12i -24 i+48i^2 = -6+12i -24 i-48= \color{red}{-6-48}\color{blue}{+12i -24i}=\color{red}{-54}\color{blue}{-12i}\)
  6. \((-6+2i) \cdot (-2+8i)= 12-48i -4 i+16i^2 = 12-48i -4 i-16= \color{red}{12-16}\color{blue}{-48i -4i}=\color{red}{-4}\color{blue}{-52i}\)
  7. \((1+4i)\cdot (-3i)= -3 i-12i^2 = \color{red}{12}\color{blue}{-3i}\)
  8. \(\frac{8+9i}{3-6i}= \frac{8+9i}{3-6i} \cdot \frac{3+6i}{3+6i} = \frac{24+48i +27 i+54i^2 }{(3)^2-(-6i)^2} = \frac{24+48i +27 i-54}{9 + 36} = \frac{-30+75i }{45} = \frac{-2}{3} - \frac{-5}{3}i \)
  9. \((8+2i)\cdot (-2i)= -16 i-4i^2 = \color{red}{4}\color{blue}{-16i}\)
  10. \((2+10i)\cdot (+6i)= +12 i+60i^2 = \color{red}{-60}\color{blue}{+12i}\)
  11. \((+9i) \cdot (-10-5i)= -90 i-45i^2 = \color{red}{45}\color{blue}{-90i}\)
  12. \((-3i) \cdot (-5-i)= +15 i+3i^2 = \color{red}{-3}\color{blue}{+15i}\)
Oefeningengenerator wiskundeoefeningen.be 2026-04-11 21:47:14
Een site van Busleyden Atheneum Mechelen