Vermenigvuldigen en delen (a+bi)

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Bereken

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

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Bereken

Verbetersleutel

1. \(\frac{8+8i}{6+i}= \frac{8+8i}{6+i} \cdot \frac{6-i}{6-i} = \frac{48-8i +48 i-8i^2 }{(6)^2-(1i)^2} = \frac{48-8i +48 i+8}{36 + 1} = \frac{56+40i }{37} = \frac{56}{37} - \frac{-40}{37}i \)
2. \((6+2i) \cdot (5+9i)= 30+54i +10 i+18i^2 = 30+54i +10 i-18= \color{red}{30-18}\color{blue}{+54i +10i}=\color{red}{12}\color{blue}{+64i}\)
3. \((-3i) \cdot (-10+7i)= +30 i-21i^2 = \color{red}{21}\color{blue}{+30i}\)
4. \((5-6i) \cdot (4-6i)= 20-30i -24 i+36i^2 = 20-30i -24 i-36= \color{red}{20-36}\color{blue}{-30i -24i}=\color{red}{-16}\color{blue}{-54i}\)
5. \(\frac{10+3i}{-6-5i}= \frac{10+3i}{-6-5i} \cdot \frac{-6+5i}{-6+5i} = \frac{-60+50i -18 i+15i^2 }{(-6)^2-(-5i)^2} = \frac{-60+50i -18 i-15}{36 + 25} = \frac{-75+32i }{61} = \frac{-75}{61} - \frac{-32}{61}i \)
6. \(\frac{-9-7i}{7+5i}= \frac{-9-7i}{7+5i} \cdot \frac{7-5i}{7-5i} = \frac{-63+45i -49 i+35i^2 }{(7)^2-(5i)^2} = \frac{-63+45i -49 i-35}{49 + 25} = \frac{-98-4i }{74} = \frac{-49}{37} + \frac{-2}{37}i \)
7. \(\frac{-7-7i}{6-8i}= \frac{-7-7i}{6-8i} \cdot \frac{6+8i}{6+8i} = \frac{-42-56i -42 i-56i^2 }{(6)^2-(-8i)^2} = \frac{-42-56i -42 i+56}{36 + 64} = \frac{14-98i }{100} = \frac{7}{50} + \frac{-49}{50}i \)
8. \(\frac{-10+9i}{-1+9i}= \frac{-10+9i}{-1+9i} \cdot \frac{-1-9i}{-1-9i} = \frac{10+90i -9 i-81i^2 }{(-1)^2-(9i)^2} = \frac{10+90i -9 i+81}{1 + 81} = \frac{91+81i }{82} = \frac{91}{82} - \frac{-81}{82}i \)
9. \(\frac{-4+5i}{-6+8i}= \frac{-4+5i}{-6+8i} \cdot \frac{-6-8i}{-6-8i} = \frac{24+32i -30 i-40i^2 }{(-6)^2-(8i)^2} = \frac{24+32i -30 i+40}{36 + 64} = \frac{64+2i }{100} = \frac{16}{25} - \frac{-1}{50}i \)
10. \(\frac{-1+8i}{9+4i}= \frac{-1+8i}{9+4i} \cdot \frac{9-4i}{9-4i} = \frac{-9+4i +72 i-32i^2 }{(9)^2-(4i)^2} = \frac{-9+4i +72 i+32}{81 + 16} = \frac{23+76i }{97} = \frac{23}{97} - \frac{-76}{97}i \)
11. \((10+2i)\cdot (-i)= -10 i-2i^2 = \color{red}{2}\color{blue}{-10i}\)
12. \((-2i) \cdot (6-i)= -12 i+2i^2 = \color{red}{-2}\color{blue}{-12i}\)