Vermenigvuldigen en delen (a+bi)

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Bereken

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

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Bereken

Verbetersleutel

1. \(\frac{-8+4i}{8+5i}= \frac{-8+4i}{8+5i} \cdot \frac{8-5i}{8-5i} = \frac{-64+40i +32 i-20i^2 }{(8)^2-(5i)^2} = \frac{-64+40i +32 i+20}{64 + 25} = \frac{-44+72i }{89} = \frac{-44}{89} - \frac{-72}{89}i \)
2. \((2-10i) \cdot (-3-9i)= -6-18i +30 i+90i^2 = -6-18i +30 i-90= \color{red}{-6-90}\color{blue}{-18i +30i}=\color{red}{-96}\color{blue}{+12i}\)
3. \(\frac{10-6i}{-8+5i}= \frac{10-6i}{-8+5i} \cdot \frac{-8-5i}{-8-5i} = \frac{-80-50i +48 i+30i^2 }{(-8)^2-(5i)^2} = \frac{-80-50i +48 i-30}{64 + 25} = \frac{-110-2i }{89} = \frac{-110}{89} + \frac{-2}{89}i \)
4. \((-10i) \cdot (-6+2i)= +60 i-20i^2 = \color{red}{20}\color{blue}{+60i}\)
5. \(\frac{-3+8i}{-3-7i}= \frac{-3+8i}{-3-7i} \cdot \frac{-3+7i}{-3+7i} = \frac{9-21i -24 i+56i^2 }{(-3)^2-(-7i)^2} = \frac{9-21i -24 i-56}{9 + 49} = \frac{-47-45i }{58} = \frac{-47}{58} + \frac{-45}{58}i \)
6. \((-2+9i) \cdot (8-4i)= -16+8i +72 i-36i^2 = -16+8i +72 i+36= \color{red}{-16+36}\color{blue}{+8i +72i}=\color{red}{20}\color{blue}{+80i}\)
7. \((2+7i)\cdot (-2i)= -4 i-14i^2 = \color{red}{14}\color{blue}{-4i}\)
8. \((10-7i) \cdot (-9+i)= -90+10i +63 i-7i^2 = -90+10i +63 i+7= \color{red}{-90+7}\color{blue}{+10i +63i}=\color{red}{-83}\color{blue}{+73i}\)
9. \((10+8i)\cdot (-3i)= -30 i-24i^2 = \color{red}{24}\color{blue}{-30i}\)
10. \((-4i) \cdot (6+6i)= -24 i-24i^2 = \color{red}{24}\color{blue}{-24i}\)
11. \((8+8i) \cdot (-3+6i)= -24+48i -24 i+48i^2 = -24+48i -24 i-48= \color{red}{-24-48}\color{blue}{+48i -24i}=\color{red}{-72}\color{blue}{+24i}\)
12. \((-8-5i) \cdot (4-5i)= -32+40i -20 i+25i^2 = -32+40i -20 i-25= \color{red}{-32-25}\color{blue}{+40i -20i}=\color{red}{-57}\color{blue}{+20i}\)