Vermenigvuldigen en delen (a+bi)

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Bereken

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

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Bereken

Verbetersleutel

1. \((-4-4i)\cdot (+4i)= -16 i-16i^2 = \color{red}{16}\color{blue}{-16i}\)
2. \(\frac{-4-5i}{-2+10i}= \frac{-4-5i}{-2+10i} \cdot \frac{-2-10i}{-2-10i} = \frac{8+40i +10 i+50i^2 }{(-2)^2-(10i)^2} = \frac{8+40i +10 i-50}{4 + 100} = \frac{-42+50i }{104} = \frac{-21}{52} - \frac{-25}{52}i \)
3. \((-6-9i) \cdot (1-3i)= -6+18i -9 i+27i^2 = -6+18i -9 i-27= \color{red}{-6-27}\color{blue}{+18i -9i}=\color{red}{-33}\color{blue}{+9i}\)
4. \(\frac{-7-4i}{3+10i}= \frac{-7-4i}{3+10i} \cdot \frac{3-10i}{3-10i} = \frac{-21+70i -12 i+40i^2 }{(3)^2-(10i)^2} = \frac{-21+70i -12 i-40}{9 + 100} = \frac{-61+58i }{109} = \frac{-61}{109} - \frac{-58}{109}i \)
5. \((-10+7i) \cdot (2-8i)= -20+80i +14 i-56i^2 = -20+80i +14 i+56= \color{red}{-20+56}\color{blue}{+80i +14i}=\color{red}{36}\color{blue}{+94i}\)
6. \((-2-8i) \cdot (6-7i)= -12+14i -48 i+56i^2 = -12+14i -48 i-56= \color{red}{-12-56}\color{blue}{+14i -48i}=\color{red}{-68}\color{blue}{-34i}\)
7. \(\frac{6-3i}{2+10i}= \frac{6-3i}{2+10i} \cdot \frac{2-10i}{2-10i} = \frac{12-60i -6 i+30i^2 }{(2)^2-(10i)^2} = \frac{12-60i -6 i-30}{4 + 100} = \frac{-18-66i }{104} = \frac{-9}{52} + \frac{-33}{52}i \)
8. \((-7-7i)\cdot (+5i)= -35 i-35i^2 = \color{red}{35}\color{blue}{-35i}\)
9. \(\frac{9-5i}{-4-i}= \frac{9-5i}{-4-i} \cdot \frac{-4+i}{-4+i} = \frac{-36+9i +20 i-5i^2 }{(-4)^2-(-1i)^2} = \frac{-36+9i +20 i+5}{16 + 1} = \frac{-31+29i }{17} = \frac{-31}{17} - \frac{-29}{17}i \)
10. \((6-2i) \cdot (4-4i)= 24-24i -8 i+8i^2 = 24-24i -8 i-8= \color{red}{24-8}\color{blue}{-24i -8i}=\color{red}{16}\color{blue}{-32i}\)
11. \((-8+9i) \cdot (2-6i)= -16+48i +18 i-54i^2 = -16+48i +18 i+54= \color{red}{-16+54}\color{blue}{+48i +18i}=\color{red}{38}\color{blue}{+66i}\)
12. \(\frac{10-3i}{1+6i}= \frac{10-3i}{1+6i} \cdot \frac{1-6i}{1-6i} = \frac{10-60i -3 i+18i^2 }{(1)^2-(6i)^2} = \frac{10-60i -3 i-18}{1 + 36} = \frac{-8-63i }{37} = \frac{-8}{37} + \frac{-63}{37}i \)