Vermenigvuldigen en delen (a+bi)

Hoofdmenu Eentje per keer  🖨

Bereken

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

---

Bereken

Verbetersleutel

1. \(\frac{-10-2i}{8+6i}= \frac{-10-2i}{8+6i} \cdot \frac{8-6i}{8-6i} = \frac{-80+60i -16 i+12i^2 }{(8)^2-(6i)^2} = \frac{-80+60i -16 i-12}{64 + 36} = \frac{-92+44i }{100} = \frac{-23}{25} - \frac{-11}{25}i \)
2. \(\frac{6+7i}{3+5i}= \frac{6+7i}{3+5i} \cdot \frac{3-5i}{3-5i} = \frac{18-30i +21 i-35i^2 }{(3)^2-(5i)^2} = \frac{18-30i +21 i+35}{9 + 25} = \frac{53-9i }{34} = \frac{53}{34} + \frac{-9}{34}i \)
3. \((1-i) \cdot (9-i)= 9-i -9 i+i^2 = 9-i -9 i-= \color{red}{9-1}\color{blue}{-i -9i}=\color{red}{8}\color{blue}{-10i}\)
4. \((-10+7i) \cdot (-6-6i)= 60+60i -42 i-42i^2 = 60+60i -42 i+42= \color{red}{60+42}\color{blue}{+60i -42i}=\color{red}{102}\color{blue}{+18i}\)
5. \((-10i) \cdot (-8+4i)= +80 i-40i^2 = \color{red}{40}\color{blue}{+80i}\)
6. \(\frac{7-i}{8-6i}= \frac{7-i}{8-6i} \cdot \frac{8+6i}{8+6i} = \frac{56+42i -8 i-6i^2 }{(8)^2-(-6i)^2} = \frac{56+42i -8 i+6}{64 + 36} = \frac{62+34i }{100} = \frac{31}{50} - \frac{-17}{50}i \)
7. \((7+5i)\cdot (+6i)= +42 i+30i^2 = \color{red}{-30}\color{blue}{+42i}\)
8. \((-5i) \cdot (-2-5i)= +10 i+25i^2 = \color{red}{-25}\color{blue}{+10i}\)
9. \((-4i) \cdot (7-5i)= -28 i+20i^2 = \color{red}{-20}\color{blue}{-28i}\)
10. \((-5i) \cdot (1+10i)= -5 i-50i^2 = \color{red}{50}\color{blue}{-5i}\)
11. \(\frac{-6-i}{5+10i}= \frac{-6-i}{5+10i} \cdot \frac{5-10i}{5-10i} = \frac{-30+60i -5 i+10i^2 }{(5)^2-(10i)^2} = \frac{-30+60i -5 i-10}{25 + 100} = \frac{-40+55i }{125} = \frac{-8}{25} - \frac{-11}{25}i \)
12. \((-5i) \cdot (-8+i)= +40 i-5i^2 = \color{red}{5}\color{blue}{+40i}\)