Complexe getallen > Optellen en aftrekkenAntwoorden van de opgaven
Teken een `xy` -assenstelsel en beide vectoren vanuit `O(0, 0)` .
Hopelijk kun je twee vectoren optellen met de parallellogrammethode. Bekijk anders
de
Dat wordt de vector `((text(-)3),(text(-)1))` vanuit `O` . Dus het complexe getal `text(-)3 + text(-)text(i)` .
`z_1 - z_2 = 1 + 2text(i) - (3 + text(i)) = 1 + 2text(i) + text(-)3 + text(-)text(i)
= text(-)2 + text(i)`
.
Ga in je figuur na dat dit klopt.
`2 + 3text(i) + 2 - text(i) = 2 + 2 + (3 - 1)text(i) = 4 + 2text(i)` .
`2 + 3text(i) - (2 - text(i)) = 2 - 2 + (3 + 1)text(i) = 0 + 4text(i) = 4text(i)` .
`3(3 + 4text(i)) + 2(5 + 2text(i)) = 9 + 12text(i) + 10 + 4text(i) = 19 + 16text(i)` .
`3 + 4text(i) - (5 + 2text(i)) = 3 + 4text(i) - 5 - 2text(i) = text(-)1 + 2text(i)` .
`text(-)2(3 + 4text(i)) + 3(5 + 2text(i)) = text(-)6 - 8text(i) + 15 + 6text(i) = 9 - 2text(i)` .
`2(3 + 4text(i)) - 3(5 + 2text(i)) = 6 + 8text(i) - 15 - 6text(i) = text(-)9 + 2text(i)` .
| `5 + 4text(i) + 3z` | `=` | `9 - z` |
beide zijden `+ z` |
| `5 + 4text(i) + 4z` | `=` | `9` |
beide zijden `- 5 - 4text(i)` |
| `4z` | `=` | `4 - 4text(i)` |
beide zijden `// 4` |
| `z` | `=` | `1 - text(i)` |
| `5(z - 2)` | `=` | `10 + 5text(i)` |
beide zijden `// 5` |
| `z - 2` | `=` | `2 + text(i)` |
beide zijden `+ 2` |
| `z` | `=` | `4 + text(i)` |
`z_1 + z_2 = 3 + 2text(i)`
`z_1 + z_2 = 22 + 20text(i)`
`z_1 - z_2 = text(-)3 + 6text(i)`
`z_1 - z_2 = text(-)8 - 4text(i)`
`z_1 + z_2 = 16 - 4text(i)` .
`z_1 - z_2 = 4 + 12text(i)` .
`4z_1 + 5z_2 = 40 + 16text(i) + 30 - 40text(i) = 70 - 24text(i)` .
`2z_1 - 3z_2 + 2text(i) = 20 + 8text(i) - (18 - 24text(i)) + 2text(i) = 2 + 34text(i)` .
`z_1 = 10cos(30^@) + 10sin(30^@)*text(i) ~~ 8,66 + 5text(i)`
`z_2 = 8cos(45^@) + 8sin(45^@)*text(i) ~~ 5,66 + 5,66text(i)`
`z_1 + z_2 ~~ 14,32 + 10,66text(i)` .
De lengte is `|z_1 + z_2| = sqrt(14,32^2 + 10,66^2) ~~ 17,85` .
Omdat `tan(varphi) ~~ (10,66)/(14,32) ~~ 0,744` is `varphi ~~ 36,7^@` .
| `3z + 5 - 2text(i)` | `=` | `text(i) - 2z` |
beide zijden `+ 2z` |
| `5z + 5 - 2text(i)` | `=` | `text(i)` |
beide zijden `- 5 + 2text(i)` |
| `5z` | `=` | `text(-)5 + 3text(i)` |
beide zijden `// 5` |
| `z` | `=` | `text(-)1 + 0,6text(i)` |
| `4(z + 2text(i))` | `=` | `12 - z` |
haakjes wegwerken |
| `4z + 8text(i)` | `=` | `12 - z` |
beide zijden `+ z` |
| `5z + 8text(i)` | `=` | `12` |
beide zijden `- 8text(i)` |
| `5z` | `=` | `12 - 8text(i)` |
beide zijden `// 5` |
| `z` | `=` | `2,4 - 1,6text(i)` |
Spoel: `z_s = 20 + 15text(j)` Ω.
Weerstand: `z_w = 12 + 0text(j) = 12` Ω.
`z = 20 + 15text(j) + 12 = 32 + 15text(j)` Ω.
Maak een `R,X` -assenstelsel (schaalbreedte `10` ), teken de vector die bij de spoel hoort en de vector van de gewone weerstand (horizontaal). En teken met de parallellogramconstructie de somvector.
Spoel: `z_s = 20 + 15text(j)` Ω.
Condensator: `z_c = 12 - 10text(j)` Ω.
`z = 20 + 15text(j) + 12 - 10txt(j) = 32 + 5text(j)` Ω.
Maak een `R,X` -assenstelsel (schaalbreedte `10` ), teken de vector die bij de spoel hoort en de vector bij de condensator. En teken met de parallellogramconstructie de somvector.
Condensator: `z_c = 15 - 12text(j)` Ω.
Weerstand: `z_w = 20` Ω.
Spoel: `z_s = 42 + 8text(j) - 20 - (15 - 12text(j) = 7 + 20text(j)` Ω.
Dus `R_s = 7` Ω en `X_(text(L)) = 20` Ω.
`z_1 + z_2 = 12 + 12text(i)`
`z_1 - z_2 = text(-)12 - 2text(i)`
`6 + 2z_1 - 3z_2 = text(-)30 - 11text(i)`
`5z_1 - 2z_2 = text(-)24 + 11text(i)`