Basisbewerkingen gemengd (a+bi)

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Bereken

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

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Bereken

Verbetersleutel

1. \((-2+2i) \cdot (-6-9i)= 12+18i -12 i-18i^2 = 12+18i -12 i+18= \color{red}{12+18}\color{blue}{+18i -12i}=\color{red}{30}\color{blue}{+6i}\)
2. \(\frac{-9+6i}{-2-6i}= \frac{-9+6i}{-2-6i} \cdot \frac{-2+6i}{-2+6i} = \frac{18-54i -12 i+36i^2 }{(-2)^2-(-6i)^2} = \frac{18-54i -12 i-36}{4 + 36} = \frac{-18-66i }{40} = \frac{-9}{20} + \frac{-33}{20}i \)
3. \(\frac{-5+5i}{5+2i}= \frac{-5+5i}{5+2i} \cdot \frac{5-2i}{5-2i} = \frac{-25+10i +25 i-10i^2 }{(5)^2-(2i)^2} = \frac{-25+10i +25 i+10}{25 + 4} = \frac{-15+35i }{29} = \frac{-15}{29} - \frac{-35}{29}i \)
4. \((9+4i)\cdot (+8i)= +72 i+32i^2 = \color{red}{-32}\color{blue}{+72i}\)
5. \((-5-5i)-(-7+4i)= -5-5i +7-4i =\color{red}{-5+7}\color{blue}{-5i -4i}=\color{red}{2}\color{blue}{-9i}\)
6. \((-6-8i)\cdot (+8i)= -48 i-64i^2 = \color{red}{64}\color{blue}{-48i}\)
7. \((7+10i)-(-5+9i)= 7+10i +5-9i =\color{red}{7+5}\color{blue}{+10i -9i}=\color{red}{12}\color{blue}{+i}\)
8. \((7+6i) \cdot (-7-6i)= -49-42i -42 i-36i^2 = -49-42i -42 i+36= \color{red}{-49+36}\color{blue}{-42i -42i}=\color{red}{-13}\color{blue}{-84i}\)
9. \(\frac{-3+4i}{5+3i}= \frac{-3+4i}{5+3i} \cdot \frac{5-3i}{5-3i} = \frac{-15+9i +20 i-12i^2 }{(5)^2-(3i)^2} = \frac{-15+9i +20 i+12}{25 + 9} = \frac{-3+29i }{34} = \frac{-3}{34} - \frac{-29}{34}i \)
10. \((8-6i) \cdot (8-8i)= 64-64i -48 i+48i^2 = 64-64i -48 i-48= \color{red}{64-48}\color{blue}{-64i -48i}=\color{red}{16}\color{blue}{-112i}\)
11. \((-5+8i)+(-7+8i)= -5+8i -7+8i =\color{red}{-5-7}\color{blue}{+8i +8i}=\color{red}{-12}\color{blue}{+16i}\)
12. \((7+9i) \cdot (-4+i)= -28+7i -36 i+9i^2 = -28+7i -36 i-9= \color{red}{-28-9}\color{blue}{+7i -36i}=\color{red}{-37}\color{blue}{-29i}\)