Vermenigvuldigen en delen (a+bi)

Hoofdmenu Eentje per keer  🖨

Bereken

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

---

Bereken

Verbetersleutel

1. \(\frac{5-3i}{-10-3i}= \frac{5-3i}{-10-3i} \cdot \frac{-10+3i}{-10+3i} = \frac{-50+15i +30 i-9i^2 }{(-10)^2-(-3i)^2} = \frac{-50+15i +30 i+9}{100 + 9} = \frac{-41+45i }{109} = \frac{-41}{109} - \frac{-45}{109}i \)
2. \((+9i) \cdot (6-10i)= +54 i-90i^2 = \color{red}{90}\color{blue}{+54i}\)
3. \(\frac{-3+4i}{2-10i}= \frac{-3+4i}{2-10i} \cdot \frac{2+10i}{2+10i} = \frac{-6-30i +8 i+40i^2 }{(2)^2-(-10i)^2} = \frac{-6-30i +8 i-40}{4 + 100} = \frac{-46-22i }{104} = \frac{-23}{52} + \frac{-11}{52}i \)
4. \((+5i) \cdot (3-8i)= +15 i-40i^2 = \color{red}{40}\color{blue}{+15i}\)
5. \((1-8i) \cdot (9+6i)= 9+6i -72 i-48i^2 = 9+6i -72 i+48= \color{red}{9+48}\color{blue}{+6i -72i}=\color{red}{57}\color{blue}{-66i}\)
6. \(\frac{-7+3i}{5+4i}= \frac{-7+3i}{5+4i} \cdot \frac{5-4i}{5-4i} = \frac{-35+28i +15 i-12i^2 }{(5)^2-(4i)^2} = \frac{-35+28i +15 i+12}{25 + 16} = \frac{-23+43i }{41} = \frac{-23}{41} - \frac{-43}{41}i \)
7. \((3-2i) \cdot (7+8i)= 21+24i -14 i-16i^2 = 21+24i -14 i+16= \color{red}{21+16}\color{blue}{+24i -14i}=\color{red}{37}\color{blue}{+10i}\)
8. \((+3i) \cdot (-6+4i)= -18 i+12i^2 = \color{red}{-12}\color{blue}{-18i}\)
9. \(\frac{-9-4i}{-7-5i}= \frac{-9-4i}{-7-5i} \cdot \frac{-7+5i}{-7+5i} = \frac{63-45i +28 i-20i^2 }{(-7)^2-(-5i)^2} = \frac{63-45i +28 i+20}{49 + 25} = \frac{83-17i }{74} = \frac{83}{74} + \frac{-17}{74}i \)
10. \(\frac{4-i}{1+7i}= \frac{4-i}{1+7i} \cdot \frac{1-7i}{1-7i} = \frac{4-28i -1 i+7i^2 }{(1)^2-(7i)^2} = \frac{4-28i -1 i-7}{1 + 49} = \frac{-3-29i }{50} = \frac{-3}{50} + \frac{-29}{50}i \)
11. \(\frac{-6+3i}{5+3i}= \frac{-6+3i}{5+3i} \cdot \frac{5-3i}{5-3i} = \frac{-30+18i +15 i-9i^2 }{(5)^2-(3i)^2} = \frac{-30+18i +15 i+9}{25 + 9} = \frac{-21+33i }{34} = \frac{-21}{34} - \frac{-33}{34}i \)
12. \(\frac{4+2i}{6-4i}= \frac{4+2i}{6-4i} \cdot \frac{6+4i}{6+4i} = \frac{24+16i +12 i+8i^2 }{(6)^2-(-4i)^2} = \frac{24+16i +12 i-8}{36 + 16} = \frac{16+28i }{52} = \frac{4}{13} - \frac{-7}{13}i \)