Vermenigvuldigen en delen (a+bi)

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Bereken

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

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Bereken

Verbetersleutel

1. \((-3-3i) \cdot (10+6i)= -30-18i -30 i-18i^2 = -30-18i -30 i+18= \color{red}{-30+18}\color{blue}{-18i -30i}=\color{red}{-12}\color{blue}{-48i}\)
2. \((6-3i) \cdot (1+i)= 6+6i -3 i-3i^2 = 6+6i -3 i+3= \color{red}{6+3}\color{blue}{+6i -3i}=\color{red}{9}\color{blue}{+3i}\)
3. \((-5-4i) \cdot (-10+8i)= 50-40i +40 i-32i^2 = 50-40i +40 i+32= \color{red}{50+32}\color{blue}{-40i +40i}=\color{red}{82}\)
4. \(\frac{-7+4i}{6+3i}= \frac{-7+4i}{6+3i} \cdot \frac{6-3i}{6-3i} = \frac{-42+21i +24 i-12i^2 }{(6)^2-(3i)^2} = \frac{-42+21i +24 i+12}{36 + 9} = \frac{-30+45i }{45} = \frac{-2}{3} - -1i\)
5. \((-10i) \cdot (6+9i)= -60 i-90i^2 = \color{red}{90}\color{blue}{-60i}\)
6. \((-7+10i)\cdot (+10i)= -70 i+100i^2 = \color{red}{-100}\color{blue}{-70i}\)
7. \((+9i) \cdot (-4+i)= -36 i+9i^2 = \color{red}{-9}\color{blue}{-36i}\)
8. \((3-8i) \cdot (10-6i)= 30-18i -80 i+48i^2 = 30-18i -80 i-48= \color{red}{30-48}\color{blue}{-18i -80i}=\color{red}{-18}\color{blue}{-98i}\)
9. \(\frac{8-5i}{-4-5i}= \frac{8-5i}{-4-5i} \cdot \frac{-4+5i}{-4+5i} = \frac{-32+40i +20 i-25i^2 }{(-4)^2-(-5i)^2} = \frac{-32+40i +20 i+25}{16 + 25} = \frac{-7+60i }{41} = \frac{-7}{41} - \frac{-60}{41}i \)
10. \(\frac{8+3i}{10+10i}= \frac{8+3i}{10+10i} \cdot \frac{10-10i}{10-10i} = \frac{80-80i +30 i-30i^2 }{(10)^2-(10i)^2} = \frac{80-80i +30 i+30}{100 + 100} = \frac{110-50i }{200} = \frac{11}{20} + \frac{-1}{4}i \)
11. \((+i) \cdot (-2-7i)= -2 i-7i^2 = \color{red}{7}\color{blue}{-2i}\)
12. \((-9+7i)\cdot (+2i)= -18 i+14i^2 = \color{red}{-14}\color{blue}{-18i}\)