Vermenigvuldigen en delen (a+bi)

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Bereken

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

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Bereken

Verbetersleutel

1. \((3+8i) \cdot (-2+i)= -6+3i -16 i+8i^2 = -6+3i -16 i-8= \color{red}{-6-8}\color{blue}{+3i -16i}=\color{red}{-14}\color{blue}{-13i}\)
2. \((9+4i)\cdot (+2i)= +18 i+8i^2 = \color{red}{-8}\color{blue}{+18i}\)
3. \((6+4i) \cdot (9+4i)= 54+24i +36 i+16i^2 = 54+24i +36 i-16= \color{red}{54-16}\color{blue}{+24i +36i}=\color{red}{38}\color{blue}{+60i}\)
4. \(\frac{6+7i}{3-7i}= \frac{6+7i}{3-7i} \cdot \frac{3+7i}{3+7i} = \frac{18+42i +21 i+49i^2 }{(3)^2-(-7i)^2} = \frac{18+42i +21 i-49}{9 + 49} = \frac{-31+63i }{58} = \frac{-31}{58} - \frac{-63}{58}i \)
5. \((10+6i) \cdot (7+10i)= 70+100i +42 i+60i^2 = 70+100i +42 i-60= \color{red}{70-60}\color{blue}{+100i +42i}=\color{red}{10}\color{blue}{+142i}\)
6. \((-4-9i) \cdot (8-3i)= -32+12i -72 i+27i^2 = -32+12i -72 i-27= \color{red}{-32-27}\color{blue}{+12i -72i}=\color{red}{-59}\color{blue}{-60i}\)
7. \((4+6i) \cdot (-10+4i)= -40+16i -60 i+24i^2 = -40+16i -60 i-24= \color{red}{-40-24}\color{blue}{+16i -60i}=\color{red}{-64}\color{blue}{-44i}\)
8. \((-8-9i) \cdot (-3+i)= 24-8i +27 i-9i^2 = 24-8i +27 i+9= \color{red}{24+9}\color{blue}{-8i +27i}=\color{red}{33}\color{blue}{+19i}\)
9. \(\frac{-10-5i}{-1-9i}= \frac{-10-5i}{-1-9i} \cdot \frac{-1+9i}{-1+9i} = \frac{10-90i +5 i-45i^2 }{(-1)^2-(-9i)^2} = \frac{10-90i +5 i+45}{1 + 81} = \frac{55-85i }{82} = \frac{55}{82} + \frac{-85}{82}i \)
10. \(\frac{-7-3i}{6+5i}= \frac{-7-3i}{6+5i} \cdot \frac{6-5i}{6-5i} = \frac{-42+35i -18 i+15i^2 }{(6)^2-(5i)^2} = \frac{-42+35i -18 i-15}{36 + 25} = \frac{-57+17i }{61} = \frac{-57}{61} - \frac{-17}{61}i \)
11. \(\frac{10-9i}{-4+2i}= \frac{10-9i}{-4+2i} \cdot \frac{-4-2i}{-4-2i} = \frac{-40-20i +36 i+18i^2 }{(-4)^2-(2i)^2} = \frac{-40-20i +36 i-18}{16 + 4} = \frac{-58+16i }{20} = \frac{-29}{10} - \frac{-4}{5}i \)
12. \((-8+2i)\cdot (-2i)= +16 i-4i^2 = \color{red}{4}\color{blue}{+16i}\)