Basisbewerkingen gemengd (a+bi)

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Bereken

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

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Bereken

Verbetersleutel

1. \((10+5i)+(-9+9i)= 10+5i -9+9i =\color{red}{10-9}\color{blue}{+5i +9i}=\color{red}{1}\color{blue}{+14i}\)
2. \(\frac{-7+4i}{-7+4i}= \frac{-7+4i}{-7+4i} \cdot \frac{-7-4i}{-7-4i} = \frac{49+28i -28 i-16i^2 }{(-7)^2-(4i)^2} = \frac{49+28i -28 i+16}{49 + 16} = \frac{65+0i }{65} = 1+ 0i\)
3. \(\frac{-6+10i}{7-4i}= \frac{-6+10i}{7-4i} \cdot \frac{7+4i}{7+4i} = \frac{-42-24i +70 i+40i^2 }{(7)^2-(-4i)^2} = \frac{-42-24i +70 i-40}{49 + 16} = \frac{-82+46i }{65} = \frac{-82}{65} - \frac{-46}{65}i \)
4. \((-4-10i)\cdot (-10i)= +40 i+100i^2 = \color{red}{-100}\color{blue}{+40i}\)
5. \((-4+10i)+(-6+3i)= -4+10i -6+3i =\color{red}{-4-6}\color{blue}{+10i +3i}=\color{red}{-10}\color{blue}{+13i}\)
6. \(\frac{6+4i}{1+6i}= \frac{6+4i}{1+6i} \cdot \frac{1-6i}{1-6i} = \frac{6-36i +4 i-24i^2 }{(1)^2-(6i)^2} = \frac{6-36i +4 i+24}{1 + 36} = \frac{30-32i }{37} = \frac{30}{37} + \frac{-32}{37}i \)
7. \(\frac{-10+4i}{-6-2i}= \frac{-10+4i}{-6-2i} \cdot \frac{-6+2i}{-6+2i} = \frac{60-20i -24 i+8i^2 }{(-6)^2-(-2i)^2} = \frac{60-20i -24 i-8}{36 + 4} = \frac{52-44i }{40} = \frac{13}{10} + \frac{-11}{10}i \)
8. \((-8-10i)-(5+5i)= -8-10i -5-5i =\color{red}{-8-5}\color{blue}{-10i -5i}=\color{red}{-13}\color{blue}{-15i}\)
9. \(\frac{-9-9i}{2+10i}= \frac{-9-9i}{2+10i} \cdot \frac{2-10i}{2-10i} = \frac{-18+90i -18 i+90i^2 }{(2)^2-(10i)^2} = \frac{-18+90i -18 i-90}{4 + 100} = \frac{-108+72i }{104} = \frac{-27}{26} - \frac{-9}{13}i \)
10. \((-4+6i)+(-1+9i)= -4+6i -1+9i =\color{red}{-4-1}\color{blue}{+6i +9i}=\color{red}{-5}\color{blue}{+15i}\)
11. \((-6+5i)\cdot (+3i)= -18 i+15i^2 = \color{red}{-15}\color{blue}{-18i}\)
12. \((-2-9i)\cdot (-5i)= +10 i+45i^2 = \color{red}{-45}\color{blue}{+10i}\)