Complexe getallen > Vermenigvuldigen en delenAntwoorden van de opgaven
`text(i)*text(i) = text(i)^2 = text(-)1` , want zo is `text(i)` immers ingevoerd. Bekijk eventueel het begin van complexe getallen nog eens.
`z_1 * z_2 = (1 + 2text(i))*(3 + text(i))` en dan de haakjes wegwerken.
`z_1 * z_2 = 1*3 + 1*text(i) + 2text(i)*3 + 2text(i)*text(i) = 3 + 7text(i) + 2text(i)^2 = 1 + 7text(i)` .
Bekijk eventueel de
Bekijk de
Probeer dit een keer te doen zonder de uitleg te bekijken.
`z_1 * z_2 = (5 + 4text(i))*(2 + 3text(i)) = 10 + 15text(i) + 8text(i) + 12text(i)^2 = text(-)2 + 23text(i)`
`z_3 * z_2 = (5 - 4text(i))*(2 + 3text(i)) = 10 + 15text(i) - 8text(i) - 12text(i)^2 = 22 + 7text(i)`
`text(i) * z_1 = text(i)*(5 - 4text(i)) = 5text(i) - 4text(i)^2 = 4 + 5text(i)`
`z_3^2 = (5 - 4text(i))*(5 - 4text(i)) = 25 - 20text(i) - 20text(i) + 16text(i)^2 = 41 - 40text(i)`
Boven de breukstreep is `(1 + 2text(i))(3 - text(i)) = 3 - text(i) + 6text(i) - 2text(i)^2 = 3 + 5text(i) + 2 = 5 + 5text(i)` .
`(z_1)/(z_2) = (5 + 4text(i))/(2 + 3text(i)) = (5 + 4text(i))/(2 + 3text(i)) * (2 - 3text(i))/(2 - 3text(i)) = (22 - 7text(i))/13 = 22/13 - 7/13 text(i)`
`(z_3)/(z_2) = (5 - 4text(i))/(2 + 3text(i)) = (5 - 4text(i))/(2 + 3text(i)) * (2 - 3text(i))/(2 - 3text(i)) = (text(-)2 - 23text(i))/13 = text(-)2/13 - 23/13 text(i)`
`1/(z_3) = 1/(5 - 4text(i)) = 1/(5 - 4text(i)) * (5 + 4text(i))/(5 + 4text(i)) = (5 + 4text(i))/31 = 5/31 + 4/31 text(i)`
`(3(3 + 4text(i)))*(2(6 + 2text(i))) = (9 + 12text(i))*(12 + 4text(i)) = 108 + 36text(i) + 144text(i) - 48 = 60 + 180text(i)` .
`(9 + 12text(i))*(12 + 4text(i)) * (12 - 4text(i))*(12 - 4text(i)) = (156 + 108text(i))160 = 0,975 + 0,675text(i)` .
`3text(i)*(3 + 4text(i))) = 9text(i) + 12text(i)^2 = text(-)12 + 9text(i)` .
`(text(i))/(6 + 2text(i)) = (text(i))/(6 + 2text(i)) * (6 - 2text(i))*(6 - 2text(i)) = (6text(i) + 2)/40 = 0,05 + 0,15 text(i)` .
`(4 text(i))/(text(-)1 + 3text(i)) = (4 text(i))/(text(-)1 + 3text(i)) * (text(-)1 - 3text(i))/(text(-)1 - 3text(i)) = (text(-)4text(i) + 12)/(10) = 1,2 - 0,4text(i)`
| `(5 + 4text(i))z` | `=` | `9 - z` |
beide zijden `+ z` |
| `(6 + 4text(i))z` | `=` | `9` |
beide zijden delen door `6 + 4text(i)` |
| `z` | `=` | `27/26 - 9/13 text(i)` |
| `5text(i)(z - 2)` | `=` | `10 + 5text(i)` |
beide zijden `// 5text(i)` |
| `z - 2` | `=` | `(10 + 5text(i))/(5 text(i)) = 1 - 2text(i)` |
beide zijden `+ 2` |
| `z` | `=` | `3 - 2text(i)` |
`z_1 * z_2 = 8 + 12text(i)`
`z_1 + z_2 = 144 + 304text(i)`
`(z_1)/(z_2) = text(-)8/13 + 12/13 text(i)`
`(z_1)/(z_2) = 21/34 + 1/34 text(i)`
`z_1 * z_2 = 92 - 56text(i)`
`(z_1)/(z_2) = 0,28 + 1,04text(i)`
`4z_1 * 5z_2 = 4 * 5 * z_1 * z_2 = 1840 - 1120text(i)`
`(2z_1)/(3z_2 + 2text(i)) = (20 + 8text(i))/(18 - 22text(i)) = 23/101 + 73/101 text(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 ~~ 20,71 + 77,27text(i)` .
De lengte is `|z_1 * z_2| = sqrt(20,71^2 + 77,27^2) ~~ 80,00` .
Omdat `tan(varphi) ~~ (77,27)/(20,71) ~~ 3,731` is `varphi ~~ 75^@` .
De lengte van `z_1 * z_2` is gelijk aan het product van de lengtes van `z_1` en `z_2` . De draaihoek van `z_1 * z_2` is gelijk aan de som van beide afzonderlijke draaihoeken.
`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) ~~ 1,21 - 0,32text(i)` .
De lengte is `|(z_1)/(z_2)| = sqrt(1,21^2 + 0,32^2) ~~ 1,25 (= 10/8)` .
Omdat `tan(varphi) ~~ (text(-)0,32)/(1,21) ~~ text(-)0,267` is `varphi ~~ text(-)15^@` .
De lengte van `z_1 * z_2` is gelijk aan het quotiënt van de lengtes van `z_1` en `z_2` . De draaihoek van `z_1 * z_2` is gelijk aan het verschil van beide afzonderlijke draaihoeken.
| `(5 - 2text(i))z` | `=` | `text(i) - 2z` |
beide zijden `+ 2z` |
| `(7 - 2text(i))z` | `=` | `text(i)` |
beide zijden delen door `7 - 2text(i)` |
| `z` | `=` | `(text(i))/(7 - 2text(i) = (7text(i) - 2)/53 = text(-)2/53 + 7/53 text(i)` |
| `z + 2text(i)(2 - text(i))` | `=` | `(2 + text(i))z` |
links de haakjes wegwerken |
| `z + 2 + 4text(i)` | `=` | `(2 + text(i))z` |
beide zijden `- z` |
| `2 + 4text(i)` | `=` | `(1 + text(i))z` |
beide zijden delen door `1 + text(i)` |
| `z` | `=` | `(2 + 4text(i))/(1 + text(i)) = 3 + text(i)` |
Spoel: `z_s = 20 + 15text(j)` Ω.
Weerstand: `z_w = 12` Ω.
`z = ((20 + 15text(j)) * 12)/(32 + 15text(j))`
`z = (240 + 180text(j))/(32 + 15text(j)) = (10380 + 2160text(j))/1249 ~~ 8,3 + 1,7text(j)` Ω.
`|z| = sqrt((10380/1249)^2 + (2160/1249)^2) ~~ 8,5` Ω.
Spoel: `z_s = 20 + 15text(j)` Ω.
Condensator: `z_c = 12 - 10text(j)` Ω.
`z = ((20 + 15text(j)) * (12 - 10text(j)))/(32 + 5text(j))`
Bereken eerst: `(20 + 15text(j)) * (12 - 10text(j)) = 390 - 20text(j)` .
`z = (390 - 20text(j))/(32 + 5text(j)) = (12380 - 2590text(j))/1049 ~~ 11,8 - 2,5text(j)` Ω.
`|z| = sqrt((12380/1049)^2 + (2590/1049)^2) ~~ 12,1` Ω.
Condensator: `z_c = 15 - 12text(j)` Ω.
Weerstand: `z_w = 20` Ω.
Condensator en weerstand samen: `z = (20*(15 - 12text(j)))/(35 - 12text(j)) = (13380 - 4800text(j))/1369`
Spoel: `z_s = 42 + 8text(j) - (13380/1369 - 4800/1369 text(j)) = 44118/1369 + 15752/1369 text(j) ~~ 32,2 + 11,5text(j)` Ω.
Dus `R_s = 32,2` Ω en `X_(text(L)) = 11,5` Ω.
`z_1 * z_2 = text(-)35 + 60text(i)`
`(z_1)/(z_2) = 35/193 + 60/193 text(i)`
`(6 + 2z_1)*3z_2 = 6 + 486text(i)`
`(6 + 2z_1)/(3z_2) = 142/579 + 26/193 text(i)`