Vermenigvuldigen en delen (a+bi)

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Bereken

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

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Bereken

Verbetersleutel

1. \(\frac{-9-9i}{8+10i}= \frac{-9-9i}{8+10i} \cdot \frac{8-10i}{8-10i} = \frac{-72+90i -72 i+90i^2 }{(8)^2-(10i)^2} = \frac{-72+90i -72 i-90}{64 + 100} = \frac{-162+18i }{164} = \frac{-81}{82} - \frac{-9}{82}i \)
2. \((10+5i) \cdot (4-6i)= 40-60i +20 i-30i^2 = 40-60i +20 i+30= \color{red}{40+30}\color{blue}{-60i +20i}=\color{red}{70}\color{blue}{-40i}\)
3. \(\frac{4+9i}{1+3i}= \frac{4+9i}{1+3i} \cdot \frac{1-3i}{1-3i} = \frac{4-12i +9 i-27i^2 }{(1)^2-(3i)^2} = \frac{4-12i +9 i+27}{1 + 9} = \frac{31-3i }{10} = \frac{31}{10} + \frac{-3}{10}i \)
4. \((1-7i) \cdot (-10+6i)= -10+6i +70 i-42i^2 = -10+6i +70 i+42= \color{red}{-10+42}\color{blue}{+6i +70i}=\color{red}{32}\color{blue}{+76i}\)
5. \((10+10i) \cdot (-7+7i)= -70+70i -70 i+70i^2 = -70+70i -70 i-70= \color{red}{-70-70}\color{blue}{+70i -70i}=\color{red}{-140}\)
6. \((2+5i) \cdot (10+5i)= 20+10i +50 i+25i^2 = 20+10i +50 i-25= \color{red}{20-25}\color{blue}{+10i +50i}=\color{red}{-5}\color{blue}{+60i}\)
7. \((-4-9i) \cdot (1+10i)= -4-40i -9 i-90i^2 = -4-40i -9 i+90= \color{red}{-4+90}\color{blue}{-40i -9i}=\color{red}{86}\color{blue}{-49i}\)
8. \((-7-4i)\cdot (-4i)= +28 i+16i^2 = \color{red}{-16}\color{blue}{+28i}\)
9. \(\frac{-1-3i}{10-10i}= \frac{-1-3i}{10-10i} \cdot \frac{10+10i}{10+10i} = \frac{-10-10i -30 i-30i^2 }{(10)^2-(-10i)^2} = \frac{-10-10i -30 i+30}{100 + 100} = \frac{20-40i }{200} = \frac{1}{10} + \frac{-1}{5}i \)
10. \((+7i) \cdot (7+7i)= +49 i+49i^2 = \color{red}{-49}\color{blue}{+49i}\)
11. \(\frac{5+2i}{-8-5i}= \frac{5+2i}{-8-5i} \cdot \frac{-8+5i}{-8+5i} = \frac{-40+25i -16 i+10i^2 }{(-8)^2-(-5i)^2} = \frac{-40+25i -16 i-10}{64 + 25} = \frac{-50+9i }{89} = \frac{-50}{89} - \frac{-9}{89}i \)
12. \((-8-4i) \cdot (-4+5i)= 32-40i +16 i-20i^2 = 32-40i +16 i+20= \color{red}{32+20}\color{blue}{-40i +16i}=\color{red}{52}\color{blue}{-24i}\)