`x=2text(i) vv x=text(-)2text(i)`
`x=1 + text(i)sqrt(15) vv x=1 - text(i)sqrt(15)`
Gebruik eventueel GeoGebra, je kunt daarin bijvoorbeeld gewoon `2 + 3text(i)` intypen en je krijgt de juiste positie in het (2D) assenstelsel.
Op de -as.
Zie de figuur in de
Bewaar dit voor een eigen samenvatting, het is alvast een eerste begin.
`1 - 2text(i) + (3 + text(i)) = 1 - 2text(i) + 3 + text(i) = 4 - text(i)` .
Dus
`1 - 2text(i) - (3 + text(i)) = 1 - 2text(i) - 3 - text(i) = text(-)2 - 3text(i)` .
Dus
Werk met iemand samen.
Doe de berekening zonder naar het voorbeeld te kijken.
`(1 - 2text(i)) * (3 + text(i)) = 1*3 + 1*text(i) - 2text(i)*3 - 2text(i)*text(i) = 3 - 5text(i) - 2text(i)^2 = 3 - 5text(i) - 2*text(-)1 = 5-5text(i)` .
Dus
en
Doen.
`(1 - 2text(i))/(3 + text(i)) = ((1 - 2text(i))*(3 - text(i)))/((3 + text(i))*(3 - text(i))) = (3 - 7text(i) + 2text(i)^2)/(9 - text(i)^2) = (1 - 7text(i))/10 = 0,1 - 0,7text(i)` .
Dus
`text(Re)(z_1/z_2)=0,1` en `text(Im)(z_1/z_2)=text(-)0,7`
`(2 + 2text(i))^3 = (2 + 2text(i)) * (2 + 2text(i)) * (2 + 2text(i)) = 8text(i)*(2 + 2text(i)) = 16text(i) + 16text(i)^2 = text(-)16 + 16text(i)` .
Doen.
`z_1+z_2=2+2text(i)+3-4text(i)=5-2text(i)`
`3z_1=3(2+2text(i))=6+6text(i)`
`2z_1-z_2=2(2+2text(i))-(3-4text(i))=4+4text(i)-3+4text(i)=1+8 text(i)`
`z_1*z_2=(2+2text(i))(3-4text(i))=6-8text(i)+6text(i)-8text(i)^2=6-2text(i)-8*text(-)1=14-2text(i)`
`bar(z_1)=2-2text(i)`
`text(i)*bar(z_1)=text(i)(2-2text(i))=2text(i)-2text(i)^2=2text(i)-2*text(-)1=2+2text(i)`
`(z_1)/(z_2)=(2+2text(i))/(3-4text(i))=(2+2text(i))/(3-4text(i))*(3+4text(i))/(3+4text(i))=((2+2text(i))(3+4text(i)))/((3-4text(i))(3+4text(i)))=(text(-)2+14text(i))/(25)=text(-)0,08+0,56text(i)`
`z_1+z_2=2+4text(i)+(1-2text(i))=3+2text(i)`
`z_1= ((2),(4))` en `z_2 = ((1),(text(-)2))` en hieruit volgt `2z_1 + z_2 = ((2),(4))+ ((1),(text(-)2)) = ((3),(2))` .
`z_1-z_2=2+4text(i)-(1-2text(i))=2+4text(i)-1+2text(i)=1+6text(i)`
`z_1= ((2),(4))` en `z_2 = ((1),(text(-)2))` en hieruit volgt `2z_1 + z_2 = ((2),(4)) - ((1),(text(-)2)) = ((1),(6))` .
`z_1= ((2),(4))` en `z_2 = ((1),(text(-)2))` en hieruit volgt `2z_1 + z_2 = ((4),(8))+ ((1),(text(-)2)) = ((5),(6))` .
`z = 3` geeft vector: `((3),(0))` .
`text(i)*z = 3 text(i)` geeft vector: `((0),(3))` .
De vector bij `3text(i)` is even lang als bij `3` , maar precies `1/2 π` om `O` gedraaid.
`z = 3 + text(i)` geeft: `((3),(1))` .
`text(i)*z = text(-)1 + 3text(i)` geeft: `((text(-)1),(3))` .
Ja, hetzelfde verband bestaat.
`z` geeft: `((x),(y))`
`text(i)*z = text(-)y + text(i)x` geeft: `((text(-)y),(x))`
Ook hier is `1/2 pi` gedraaid om `O` .
`text(i)^2 = text(-)1` voldoet ook aan dit verband.
`text(i)^2=1*text(i)*text(i)` , je draait de vector bij `1` tweemaal over een hoek van `1/2 pi` tegen de klok in. Dan kom je bij `text(-)1` uit.
`z=2+3text(i)-(5+4text(i))=text(-)3-text(i)` en hieruit volgt `text(Re)(z)=text(-)3` en `text(Im)(z)=text(-)1` .
`z=(2+text(i))(3-2text(i))=8-text(i)` en hieruit volgt `text(Re)(z)=8` en `text(Im)(z)=text(-)1` .
`z=2+3i(2-5i)^2=2+3i(text(-)21-20i)=62-63i` en hieruit volgt `text(Re)(z)=62` en `text(Im)(z)=text(-)63` .
`z=(3+4text(i))(3-4text(i))=25` en hieruit volgt `text(Re)(z)=25` en `text(Im)(z)=0` .
`z=6+4text(i)-(1+text(i))^2-(2+3text(i))text(i)=6+4text(i)-2text(i)-2text(i)+3=9` en hieruit volgt `text(Re)(z)=9` en `text(Im)(z)=0` .
`z=(3+2text(i))/(5+3text(i))=(3+2text(i))/(5+3text(i))*(5-3text(i))/(5-3text(i))=((3+2text(i))(5-3text(i)))/((5+3text(i))(5-3text(i)))=(21+text(i))/(34)` en hieruit volgt `text(Re)(z)=21/34` en `text(Im)(z)=1/34` .
`(z-2)^2=text(-)9` geeft `z-2=text(-)3text(i) vv z-2=3text(i)` en `z=2-3text(i) vv z =2+3text(i)` .
`2(z-text(i))^2+8=0` geeft `(z-text(i))^2=text(-)4` .
En `z-text(i)=text(-)2text(i) vv z-text(i)=2text(i)` geeft `z=text(-)text(i) vv z=3text(i)` .
`12+z^2=4` geeft `z^2=text(-)8` en `z=text(-)sqrt(8)*text(i) vv z=sqrt(8)*text(i)` .
`5z+2=3z+4text(i)` geeft `2z=4text(i)-2` en `z=text(-)1+2text(i)`
`5z+2=3text(i)z+4text(i)` geeft `z*(5-3text(i))=text(-)2+4text(i)` en `z=(text(-)2+4text(i))/(5-3text(i))` .
Dus `z=(text(-)2+4text(i))/(5-3text(i))*(5+3text(i))/(5+3text(i))=(text(-)22+14text(i))/(34)=(text(-)11)/17+7/17text(i)` .
`(2-3text(i))^2` |
`=` |
`text(-)5-12text(i)` |
|
`(2-3text(i))^4` |
`=` |
`(text(-)5-12text(i))^2=text(-)119+120text(i)` |
|
`(2-3text(i))^5` |
`=` |
`(text(-)119+120text(i))(2-3text(i))=122+597text(i)` |
Dus `text(Re)((2 -3 text(i)) ^5) = 122` en `text(Im)( (2 -3 text(i)) ^5)=597`
`z_1 + z_2 = (a + c) + (b + d)text(i)`
`z_1 - z_2 = (a - c) + (b - d)text(i)`
`z_1 * z_2 = (ac - bd) + (ad + bc)text(i)`
`(z_1)/(z_2) = (ac+bd)/(c^2+d^2) + (bc-ad)/(c^2+d^2)text(i)`
,
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`z = text(-i) +- 4text(i)` , dus `z = text(-)5text(i) vv z = 3text(i)` .
`z = (5+2text(i))/(text(-)3+2text(i)) = text(-) 11/13 - 16/13 text(i)`