Vermenigvuldigen en delen (a+bi)

Hoofdmenu Eentje per keer  🖨

Bereken

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

---

Bereken

Verbetersleutel

1. \(\frac{-3-i}{-1+2i}= \frac{-3-i}{-1+2i} \cdot \frac{-1-2i}{-1-2i} = \frac{3+6i +1 i+2i^2 }{(-1)^2-(2i)^2} = \frac{3+6i +1 i-2}{1 + 4} = \frac{1+7i }{5} = \frac{1}{5} - \frac{-7}{5}i \)
2. \((+9i) \cdot (-10+9i)= -90 i+81i^2 = \color{red}{-81}\color{blue}{-90i}\)
3. \((4-8i) \cdot (2+2i)= 8+8i -16 i-16i^2 = 8+8i -16 i+16= \color{red}{8+16}\color{blue}{+8i -16i}=\color{red}{24}\color{blue}{-8i}\)
4. \(\frac{-8-5i}{3+i}= \frac{-8-5i}{3+i} \cdot \frac{3-i}{3-i} = \frac{-24+8i -15 i+5i^2 }{(3)^2-(1i)^2} = \frac{-24+8i -15 i-5}{9 + 1} = \frac{-29-7i }{10} = \frac{-29}{10} + \frac{-7}{10}i \)
5. \((-3+9i) \cdot (6+4i)= -18-12i +54 i+36i^2 = -18-12i +54 i-36= \color{red}{-18-36}\color{blue}{-12i +54i}=\color{red}{-54}\color{blue}{+42i}\)
6. \((-6+i)\cdot (-4i)= +24 i-4i^2 = \color{red}{4}\color{blue}{+24i}\)
7. \((-2+3i)\cdot (-3i)= +6 i-9i^2 = \color{red}{9}\color{blue}{+6i}\)
8. \((-3-10i) \cdot (-1+6i)= 3-18i +10 i-60i^2 = 3-18i +10 i+60= \color{red}{3+60}\color{blue}{-18i +10i}=\color{red}{63}\color{blue}{-8i}\)
9. \((-5-5i) \cdot (2+7i)= -10-35i -10 i-35i^2 = -10-35i -10 i+35= \color{red}{-10+35}\color{blue}{-35i -10i}=\color{red}{25}\color{blue}{-45i}\)
10. \(\frac{9+8i}{5+6i}= \frac{9+8i}{5+6i} \cdot \frac{5-6i}{5-6i} = \frac{45-54i +40 i-48i^2 }{(5)^2-(6i)^2} = \frac{45-54i +40 i+48}{25 + 36} = \frac{93-14i }{61} = \frac{93}{61} + \frac{-14}{61}i \)
11. \((-4-7i) \cdot (6-2i)= -24+8i -42 i+14i^2 = -24+8i -42 i-14= \color{red}{-24-14}\color{blue}{+8i -42i}=\color{red}{-38}\color{blue}{-34i}\)
12. \(\frac{-1+3i}{3+10i}= \frac{-1+3i}{3+10i} \cdot \frac{3-10i}{3-10i} = \frac{-3+10i +9 i-30i^2 }{(3)^2-(10i)^2} = \frac{-3+10i +9 i+30}{9 + 100} = \frac{27+19i }{109} = \frac{27}{109} - \frac{-19}{109}i \)