Vermenigvuldigen en delen (a+bi)

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Bereken

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

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Bereken

Verbetersleutel

1. \(\frac{-9-10i}{-10+6i}= \frac{-9-10i}{-10+6i} \cdot \frac{-10-6i}{-10-6i} = \frac{90+54i +100 i+60i^2 }{(-10)^2-(6i)^2} = \frac{90+54i +100 i-60}{100 + 36} = \frac{30+154i }{136} = \frac{15}{68} - \frac{-77}{68}i \)
2. \((-10+8i) \cdot (7+i)= -70-10i +56 i+8i^2 = -70-10i +56 i-8= \color{red}{-70-8}\color{blue}{-10i +56i}=\color{red}{-78}\color{blue}{+46i}\)
3. \((10+i) \cdot (3-3i)= 30-30i +3 i-3i^2 = 30-30i +3 i+3= \color{red}{30+3}\color{blue}{-30i +3i}=\color{red}{33}\color{blue}{-27i}\)
4. \((-8-2i)\cdot (+8i)= -64 i-16i^2 = \color{red}{16}\color{blue}{-64i}\)
5. \((3-3i)\cdot (+4i)= +12 i-12i^2 = \color{red}{12}\color{blue}{+12i}\)
6. \((-6+3i) \cdot (9-9i)= -54+54i +27 i-27i^2 = -54+54i +27 i+27= \color{red}{-54+27}\color{blue}{+54i +27i}=\color{red}{-27}\color{blue}{+81i}\)
7. \((1-3i) \cdot (-10-8i)= -10-8i +30 i+24i^2 = -10-8i +30 i-24= \color{red}{-10-24}\color{blue}{-8i +30i}=\color{red}{-34}\color{blue}{+22i}\)
8. \((-4-4i) \cdot (-6+6i)= 24-24i +24 i-24i^2 = 24-24i +24 i+24= \color{red}{24+24}\color{blue}{-24i +24i}=\color{red}{48}\)
9. \(\frac{6+3i}{-10-i}= \frac{6+3i}{-10-i} \cdot \frac{-10+i}{-10+i} = \frac{-60+6i -30 i+3i^2 }{(-10)^2-(-1i)^2} = \frac{-60+6i -30 i-3}{100 + 1} = \frac{-63-24i }{101} = \frac{-63}{101} + \frac{-24}{101}i \)
10. \((8+5i) \cdot (2-5i)= 16-40i +10 i-25i^2 = 16-40i +10 i+25= \color{red}{16+25}\color{blue}{-40i +10i}=\color{red}{41}\color{blue}{-30i}\)
11. \(\frac{-4+9i}{-10-9i}= \frac{-4+9i}{-10-9i} \cdot \frac{-10+9i}{-10+9i} = \frac{40-36i -90 i+81i^2 }{(-10)^2-(-9i)^2} = \frac{40-36i -90 i-81}{100 + 81} = \frac{-41-126i }{181} = \frac{-41}{181} + \frac{-126}{181}i \)
12. \(\frac{-1-8i}{3+i}= \frac{-1-8i}{3+i} \cdot \frac{3-i}{3-i} = \frac{-3+i -24 i+8i^2 }{(3)^2-(1i)^2} = \frac{-3+i -24 i-8}{9 + 1} = \frac{-11-23i }{10} = \frac{-11}{10} + \frac{-23}{10}i \)