Basisbewerkingen gemengd (a+bi)

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Bereken

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

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Bereken

Verbetersleutel

1. \(\frac{-10-9i}{-10+8i}= \frac{-10-9i}{-10+8i} \cdot \frac{-10-8i}{-10-8i} = \frac{100+80i +90 i+72i^2 }{(-10)^2-(8i)^2} = \frac{100+80i +90 i-72}{100 + 64} = \frac{28+170i }{164} = \frac{7}{41} - \frac{-85}{82}i \)
2. \((-9-10i)+(10-5i)= -9-10i +10-5i =\color{red}{-9+10}\color{blue}{-10i -5i}=\color{red}{1}\color{blue}{-15i}\)
3. \(\frac{-2-9i}{-10+10i}= \frac{-2-9i}{-10+10i} \cdot \frac{-10-10i}{-10-10i} = \frac{20+20i +90 i+90i^2 }{(-10)^2-(10i)^2} = \frac{20+20i +90 i-90}{100 + 100} = \frac{-70+110i }{200} = \frac{-7}{20} - \frac{-11}{20}i \)
4. \((8-8i) \cdot (-5+7i)= -40+56i +40 i-56i^2 = -40+56i +40 i+56= \color{red}{-40+56}\color{blue}{+56i +40i}=\color{red}{16}\color{blue}{+96i}\)
5. \((-3i) \cdot (-7-10i)= +21 i+30i^2 = \color{red}{-30}\color{blue}{+21i}\)
6. \(\frac{1+2i}{1+2i}= \frac{1+2i}{1+2i} \cdot \frac{1-2i}{1-2i} = \frac{1-2i +2 i-4i^2 }{(1)^2-(2i)^2} = \frac{1-2i +2 i+4}{1 + 4} = \frac{5+0i }{5} = 1+ 0i\)
7. \((-4-2i) \cdot (3-8i)= -12+32i -6 i+16i^2 = -12+32i -6 i-16= \color{red}{-12-16}\color{blue}{+32i -6i}=\color{red}{-28}\color{blue}{+26i}\)
8. \((-7+7i)\cdot (-3i)= +21 i-21i^2 = \color{red}{21}\color{blue}{+21i}\)
9. \((-9+10i) \cdot (-3-5i)= 27+45i -30 i-50i^2 = 27+45i -30 i+50= \color{red}{27+50}\color{blue}{+45i -30i}=\color{red}{77}\color{blue}{+15i}\)
10. \((2-9i) \cdot (-7-4i)= -14-8i +63 i+36i^2 = -14-8i +63 i-36= \color{red}{-14-36}\color{blue}{-8i +63i}=\color{red}{-50}\color{blue}{+55i}\)
11. \(\frac{3-i}{-10-8i}= \frac{3-i}{-10-8i} \cdot \frac{-10+8i}{-10+8i} = \frac{-30+24i +10 i-8i^2 }{(-10)^2-(-8i)^2} = \frac{-30+24i +10 i+8}{100 + 64} = \frac{-22+34i }{164} = \frac{-11}{82} - \frac{-17}{82}i \)
12. \(\frac{6-10i}{-4+4i}= \frac{6-10i}{-4+4i} \cdot \frac{-4-4i}{-4-4i} = \frac{-24-24i +40 i+40i^2 }{(-4)^2-(4i)^2} = \frac{-24-24i +40 i-40}{16 + 16} = \frac{-64+16i }{32} = -2- \frac{-1}{2}i \)