Basisbewerkingen gemengd (a+bi)

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Bereken

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

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Bereken

Verbetersleutel

1. \(\frac{10-3i}{8-i}= \frac{10-3i}{8-i} \cdot \frac{8+i}{8+i} = \frac{80+10i -24 i-3i^2 }{(8)^2-(-1i)^2} = \frac{80+10i -24 i+3}{64 + 1} = \frac{83-14i }{65} = \frac{83}{65} + \frac{-14}{65}i \)
2. \((-8+8i)+(-3-4i)= -8+8i -3-4i =\color{red}{-8-3}\color{blue}{+8i -4i}=\color{red}{-11}\color{blue}{+4i}\)
3. \(\frac{-5-10i}{-1-2i}= \frac{-5-10i}{-1-2i} \cdot \frac{-1+2i}{-1+2i} = \frac{5-10i +10 i-20i^2 }{(-1)^2-(-2i)^2} = \frac{5-10i +10 i+20}{1 + 4} = \frac{25+0i }{5} = 5+ 0i\)
4. \((8+7i)+(-10+2i)= 8+7i -10+2i =\color{red}{8-10}\color{blue}{+7i +2i}=\color{red}{-2}\color{blue}{+9i}\)
5. \((-10+5i)+(-4+6i)= -10+5i -4+6i =\color{red}{-10-4}\color{blue}{+5i +6i}=\color{red}{-14}\color{blue}{+11i}\)
6. \(\frac{4-2i}{1-3i}= \frac{4-2i}{1-3i} \cdot \frac{1+3i}{1+3i} = \frac{4+12i -2 i-6i^2 }{(1)^2-(-3i)^2} = \frac{4+12i -2 i+6}{1 + 9} = \frac{10+10i }{10} = 1- -1i\)
7. \(\frac{-2-7i}{3-8i}= \frac{-2-7i}{3-8i} \cdot \frac{3+8i}{3+8i} = \frac{-6-16i -21 i-56i^2 }{(3)^2-(-8i)^2} = \frac{-6-16i -21 i+56}{9 + 64} = \frac{50-37i }{73} = \frac{50}{73} + \frac{-37}{73}i \)
8. \((-5+4i)+(-1-4i)= -5+4i -1-4i =\color{red}{-5-1}\color{blue}{+4i -4i}=\color{red}{-6}\)
9. \((-8+8i)-(7-3i)= -8+8i -7+3i =\color{red}{-8-7}\color{blue}{+8i +3i}=\color{red}{-15}\color{blue}{+11i}\)
10. \(\frac{7+6i}{-4-4i}= \frac{7+6i}{-4-4i} \cdot \frac{-4+4i}{-4+4i} = \frac{-28+28i -24 i+24i^2 }{(-4)^2-(-4i)^2} = \frac{-28+28i -24 i-24}{16 + 16} = \frac{-52+4i }{32} = \frac{-13}{8} - \frac{-1}{8}i \)
11. \((8-i)+(3+10i)= 8-i +3+10i =\color{red}{8+3}\color{blue}{-i +10i}=\color{red}{11}\color{blue}{+9i}\)
12. \((-8+8i)+(7-5i)= -8+8i +7-5i =\color{red}{-8+7}\color{blue}{+8i -5i}=\color{red}{-1}\color{blue}{+3i}\)