Vermenigvuldigen en delen (a+bi)

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Bereken

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

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Bereken

Verbetersleutel

1. \(\frac{4+i}{-1-6i}= \frac{4+i}{-1-6i} \cdot \frac{-1+6i}{-1+6i} = \frac{-4+24i -1 i+6i^2 }{(-1)^2-(-6i)^2} = \frac{-4+24i -1 i-6}{1 + 36} = \frac{-10+23i }{37} = \frac{-10}{37} - \frac{-23}{37}i \)
2. \(\frac{-10+6i}{-6-i}= \frac{-10+6i}{-6-i} \cdot \frac{-6+i}{-6+i} = \frac{60-10i -36 i+6i^2 }{(-6)^2-(-1i)^2} = \frac{60-10i -36 i-6}{36 + 1} = \frac{54-46i }{37} = \frac{54}{37} + \frac{-46}{37}i \)
3. \((10+4i) \cdot (-10+9i)= -100+90i -40 i+36i^2 = -100+90i -40 i-36= \color{red}{-100-36}\color{blue}{+90i -40i}=\color{red}{-136}\color{blue}{+50i}\)
4. \((+7i) \cdot (-7+6i)= -49 i+42i^2 = \color{red}{-42}\color{blue}{-49i}\)
5. \(\frac{-7-7i}{3-5i}= \frac{-7-7i}{3-5i} \cdot \frac{3+5i}{3+5i} = \frac{-21-35i -21 i-35i^2 }{(3)^2-(-5i)^2} = \frac{-21-35i -21 i+35}{9 + 25} = \frac{14-56i }{34} = \frac{7}{17} + \frac{-28}{17}i \)
6. \((-1-8i) \cdot (7-10i)= -7+10i -56 i+80i^2 = -7+10i -56 i-80= \color{red}{-7-80}\color{blue}{+10i -56i}=\color{red}{-87}\color{blue}{-46i}\)
7. \(\frac{6+5i}{-8-5i}= \frac{6+5i}{-8-5i} \cdot \frac{-8+5i}{-8+5i} = \frac{-48+30i -40 i+25i^2 }{(-8)^2-(-5i)^2} = \frac{-48+30i -40 i-25}{64 + 25} = \frac{-73-10i }{89} = \frac{-73}{89} + \frac{-10}{89}i \)
8. \(\frac{-5+4i}{-4-i}= \frac{-5+4i}{-4-i} \cdot \frac{-4+i}{-4+i} = \frac{20-5i -16 i+4i^2 }{(-4)^2-(-1i)^2} = \frac{20-5i -16 i-4}{16 + 1} = \frac{16-21i }{17} = \frac{16}{17} + \frac{-21}{17}i \)
9. \(\frac{-6-4i}{-3-8i}= \frac{-6-4i}{-3-8i} \cdot \frac{-3+8i}{-3+8i} = \frac{18-48i +12 i-32i^2 }{(-3)^2-(-8i)^2} = \frac{18-48i +12 i+32}{9 + 64} = \frac{50-36i }{73} = \frac{50}{73} + \frac{-36}{73}i \)
10. \((5+5i)\cdot (-8i)= -40 i-40i^2 = \color{red}{40}\color{blue}{-40i}\)
11. \((+6i) \cdot (-3-i)= -18 i-6i^2 = \color{red}{6}\color{blue}{-18i}\)
12. \(\frac{10+3i}{8-3i}= \frac{10+3i}{8-3i} \cdot \frac{8+3i}{8+3i} = \frac{80+30i +24 i+9i^2 }{(8)^2-(-3i)^2} = \frac{80+30i +24 i-9}{64 + 9} = \frac{71+54i }{73} = \frac{71}{73} - \frac{-54}{73}i \)