Vermenigvuldigen en delen (a+bi)

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Bereken

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

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Bereken

Verbetersleutel

1. \(\frac{5+i}{9+6i}= \frac{5+i}{9+6i} \cdot \frac{9-6i}{9-6i} = \frac{45-30i +9 i-6i^2 }{(9)^2-(6i)^2} = \frac{45-30i +9 i+6}{81 + 36} = \frac{51-21i }{117} = \frac{17}{39} + \frac{-7}{39}i \)
2. \((1+9i) \cdot (3-10i)= 3-10i +27 i-90i^2 = 3-10i +27 i+90= \color{red}{3+90}\color{blue}{-10i +27i}=\color{red}{93}\color{blue}{+17i}\)
3. \(\frac{4+i}{-6+6i}= \frac{4+i}{-6+6i} \cdot \frac{-6-6i}{-6-6i} = \frac{-24-24i -6 i-6i^2 }{(-6)^2-(6i)^2} = \frac{-24-24i -6 i+6}{36 + 36} = \frac{-18-30i }{72} = \frac{-1}{4} + \frac{-5}{12}i \)
4. \((7-6i)\cdot (+4i)= +28 i-24i^2 = \color{red}{24}\color{blue}{+28i}\)
5. \((+8i) \cdot (4+10i)= +32 i+80i^2 = \color{red}{-80}\color{blue}{+32i}\)
6. \((1-3i) \cdot (4+5i)= 4+5i -12 i-15i^2 = 4+5i -12 i+15= \color{red}{4+15}\color{blue}{+5i -12i}=\color{red}{19}\color{blue}{-7i}\)
7. \((-8+3i) \cdot (6-4i)= -48+32i +18 i-12i^2 = -48+32i +18 i+12= \color{red}{-48+12}\color{blue}{+32i +18i}=\color{red}{-36}\color{blue}{+50i}\)
8. \(\frac{1-9i}{-5-10i}= \frac{1-9i}{-5-10i} \cdot \frac{-5+10i}{-5+10i} = \frac{-5+10i +45 i-90i^2 }{(-5)^2-(-10i)^2} = \frac{-5+10i +45 i+90}{25 + 100} = \frac{85+55i }{125} = \frac{17}{25} - \frac{-11}{25}i \)
9. \((-3-2i) \cdot (-1+9i)= 3-27i +2 i-18i^2 = 3-27i +2 i+18= \color{red}{3+18}\color{blue}{-27i +2i}=\color{red}{21}\color{blue}{-25i}\)
10. \((-10i) \cdot (-1-5i)= +10 i+50i^2 = \color{red}{-50}\color{blue}{+10i}\)
11. \((-7+4i)\cdot (+4i)= -28 i+16i^2 = \color{red}{-16}\color{blue}{-28i}\)
12. \(\frac{-2+2i}{-2+i}= \frac{-2+2i}{-2+i} \cdot \frac{-2-i}{-2-i} = \frac{4+2i -4 i-2i^2 }{(-2)^2-(1i)^2} = \frac{4+2i -4 i+2}{4 + 1} = \frac{6-2i }{5} = \frac{6}{5} + \frac{-2}{5}i \)