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Antwoorden van de opgaven

Opgave V1
a

`x = text(-)1 +- text(i)`

b

`x = 0 vv x = +- 4text(i)`

Opgave 1
a

`z_1 = text(e)^(1/4pi text(i))=cos(1/4pi)+text(i)sin(1/4pi)=1/2sqrt(2)+1/2sqrt(2)text(i)`
`z_2 = text(e)^(3/4pi text(i))=cos(3/4pi)+text(i)sin(3/4pi)=text(-)1/2sqrt(2)+1/2sqrt(2)text(i)`
`z_3 = text(e)^(1 1/4pi text(i))=cos(1 1/4pi)+text(i)sin(1 1/4pi)=text(-)1/2sqrt(2)-1/2sqrt(2)text(i)`
`z_4 = text(e)^(1 3/4pi text(i))=cos(1 3/4pi)+text(i)sin(1 3/4pi)=1/2sqrt(2)+1/2sqrt(2)text(i)`

b

`z^4=text(e)^(k*2pi text(i))`
`z=(text(e)^(k*2pi text(i)))^(1/4)`
`z=text(e)^(k*1/2pi text(i))`
`z=text(e)^0 = 1 vv z=text(e)^(1/2pi text(i))=i vv z=text(e)^(pi text(i))=text(-)1 vv z=text(e)^(1 1/2pi text(i))=text(-i)`

Opgave 2
a

Je deelt dan beide zijden door `1 + text(i)` .

b

`z=(2-2i)/(1+i)=(2-2i)/(1+i)*(1-i)/(1-i)=(text(-)4i)/2=text(-)2i`

b

`(1+text(i))z`

`=`

`2z-2text(i)`

`z(1+text(i))-2z`

`=`

`text(-)2text(i)`

`z(text(-)1+text(i))`

`=`

`text(-)2text(i)`

`z`

`=`

`(text(-)2text(i))/(text(-)1+text(i))`

`z`

`=`

`(text(-)2text(i))/(text(-)1+text(i))*(text(-)1-text(i))/(text(-)1-text(i))=(text(-)2+2text(i))/2=text(-)1+text(i)`

Opgave 3
a

`z^3=1=text(e)^(k*2pi text(i))` geeft `z=(text(e)^(k*2pi text(i)))^(1/3)` , dus `z=text(e)^(k*2/3pi text(i))` .

Oplossingen: `z=text(e)^0 = 1 vv z=text(e)^(2/3pi text(i)) vv z=text(e)^(4/3pi text(i))` .

Ofwel: `z=1 vv z=text(-)0,5 +0,5 text(i)sqrt(3 ) vv z=text(-)0,5 -0,5 text(i)sqrt(3 )` .

b

`z^4=text(-)81=81text(e)^((pi + k*2pi) text(i))` geeft `z=3text(e)^((1/4 pi + k*1/2 pi) text(i))` .

Oplossingen:

`z_1=3text(e)^(1/4pi text(i))=3(cos(1/4pi)+isin(1/4pi))=3(1/2sqrt(2)+1/2sqrt(2)i)=1 1/2sqrt(2)+1 1/2sqrt(2)i`

`z_2=3text(e)^(3/4pi text(i))=3(cos(3/4pi)+isin(1/4pi))=3(text(-)1/2sqrt(2)+1/2sqrt(2)i)=text(-)1 1/2sqrt(2)+1 1/2sqrt(2)i`

`z_3=3text(e)^(1 1/4pi text(i))=3(cos(1 1/4pi)+isin(1 1/4pi))=3(text(-)1/2sqrt(2)-1/2sqrt(2)i)=text(-)1 1/2sqrt(2)-1 1/2sqrt(2)i`

`z_4=3text(e)^(1 3/4pi text(i))=3(cos(1 3/4pi)+isin(1 3/4pi))=3(1/2sqrt(2)-1/2sqrt(2)i)=1 1/2sqrt(2)-1 1/2sqrt(2)i`

Opgave 4
a

Omdat `text(-)1 = text(i)^2` .

b

`(z+1)^2 = 4text(i)^2` betekent `z + 1 = 2text(i) vv z + 1 = text(-)2text(i)` .

c

`(z+1)^2 = text(-)4 = 4 text(e)^(pi text(i) + k*2pi text(i))` betekent `z + 1 = 2text(e)^(1/2 pi text(i) + k*pi text(i))` en dus `z + 1 = 2text(i) vv z + 1 = text(-)2text(i)` , enzovoorts.

d

`z^2 + 2z + 5 = 0` heeft `a=1` , `b=2` en `c=5` .

Dus `z = (text(-)2 +- sqrt(2^2-4*1*5))/(2*1) = (text(-)2 +- sqrt(text(-)16))/(2) = (text(-)2+-4text(i))/2 = text(-)1 +- 2text(i)` .

e

Kies één van de verschillende oplossingsmethoden. Je vindt:
z = - 2,5 ± 0,5 i 15 .

Opgave 5
a

`(z+1-text(i)) ^6=text(-i)=text(e)^((1 1/2 pi+k*2pi)*text(i))`

Dus `z+1-text(i)=text(e)^((1/4 pi+k*1/3 pi)*text(i))` .

Oplossing: `z= text(-)1 + text(i) + text(e)^((1/4 pi+k*1/3 pi)*text(i))` .

b

`z_1 =text(-)1 +0,5 sqrt(2 )+(1 +0,5 sqrt(2))text(i)`
`z_2 =text(-)1 +cos(7/12π)+(1 +sin(7/12π))text(i)`
`z_3 =text(-)1 +cos(11/12π)+(1 +sin(11/12π))text(i)`
`z_4 =text(-)1 -0,5 sqrt(2 )+(1 -0,5 sqrt(2))text(i)`
`z_5 =text(-)1 +cos(19/12π)+(1 +sin(19/12π))text(i)`
`z_6 =text(-)1 +cos(23/12π)+(1 +sin(23/12π))text(i)`

c

Doen.

Opgave 6
a

`2z+5text(i)`

`=`

`3text(i)z-4`

`z(2-3text(i))`

`=`

`text(-)4-5text(i)`

`z`

`=`

`(text(-)4-5text(i))/(2-3text(i))`

`z`

`=`

`7/13-22/13text(i)`

b

`(z+1)/(z-text(i))`

`=`

`2`

`z+1`

`=`

`2z-2text(i)`

`z`

`=`

`1+2text(i)`

c

`3/z`

`=`

`2text(i)`

`z`

`=`

`3/(2text(i))`

`z`

`=`

`text(-)1,5text(i)`

Opgave 7
a

`z^5=1=text(e)^(k*2pi i)` geeft `z=text(e)^(2/5k*pi text(i))`
De oplossingen zijn:
`z=text(e)^0=1 vv z=text(e)^(2/5pi text(i)) vv z=text(e)^(4/5pi text(i)) vv z=text(e)^(6/5pi text(i)) vv z=text(e)^(8/5pi text(i))`

b

`z=2text(e)^(1/3pi text(i)) vv z=2text(e)^(pi i) vv z=2text(e)^(1 2/3pi text(i))`

Opgave 8
a

`z^3=text(i)=text(e)^((1/2 pi + k*2pi)text(i))` geeft `z=text(e)^((1/6 pi + k*2/3 pi)text(i))` .

Dus `z =0,5 sqrt(3 )+0,5 text(i) vv z =text(-)0,5 sqrt(3)+0,5 text(i) vv z =text(-)text(i)` .

b

`z^4=text(-)16=16text(e)^((pi+k*2pi) text(i))` geeft `z=2text(e)^((1/4pi+k*1/2pi) text(i))` .
De oplossingen zijn:
`z_1=2text(e)^(1/4pi text(i))=2(1/2sqrt(2)+1/2sqrt(2)text(i))=sqrt(2)+sqrt(2)text(i)`
`z_2=2text(e)^(3/4pi text(i))=2(text(-)1/2sqrt(2)+1/2sqrt(2)text(i))=text(-)sqrt(2)+sqrt(2)text(i)`
`z_3=2text(e)^(5/4pi text(i))=2(text(-)1/2sqrt(2)-1/2sqrt(2)text(i))=text(-)sqrt(2)-sqrt(2)text(i)`
`z_4=2text(e)^(7/4pi text(i))=2(1/2sqrt(2)-1/2sqrt(2)text(i))=sqrt(2)-sqrt(2)text(i)`

c

`z^2=text(-i)=text(e)^((1 1/2pi+k*2pi)text(i))` geeft `z=text(e)^((3/4pi+k*pi)text(i))` .
De oplossingen zijn:
`z_1=text(e)^(3/4pi text(i))=text(-)1/2sqrt(2)+1/2sqrt(2)text(i)`
`z_2=text(e)^(1 3/4pi text(i))=1/2sqrt(2)-1/2sqrt(2)text(i)`

d

`z^3=text(-)27text(i)=27text(e)^((1 1/2pi+k*2pi)text(i))` geeft `z=3text(e)^((1/2pi+k*2/3pi)text(i))` .
De oplossingen zijn:
`z_1=3text(e)^(1/2pi text(i))=3text(i)`
`z_2=3text(e)^(7/6pi text(i))=3(text(-)1/2sqrt(3)-1/2text(i))=text(-)1 1/2sqrt(3)-1 1/2text(i)`
`z_3=3text(e)^(11/6pi text(i))=3(1/2sqrt(3)-1/2text(i))=1 1/2sqrt(3)-1 1/2text(i)`

e

`z^4=text(-)8 +8sqrt(3) text(i)=16text(e)^((2/3pi + k*2pi) text(i))` geeft `z=2text(e)^((1/6pi + k*1/2pi) text(i))` .
De oplossingen zijn:
`z_1=2text(e)^(1/6pi text(i))=2(1/2sqrt(3)+1/2text(i))=sqrt(3)+text(i)`
`z_2=2text(e)^(2/3pi text(i))=2(text(-)1/2+1/2sqrt(3)text(i))=text(-)1+sqrt(3)text(i)`
`z_3=2text(e)^(7/6pi text(i))=2(text(-)1/2sqrt(3)-1/2text(i))=text(-)sqrt(3)-text(i)`
`z_4=2text(e)^(5/3pi text(i))=2(1/2-1/2sqrt(3)text(i))=1-sqrt(3)text(i)`

f

`text(i)z^2=0,5` geeft `z^2=(0,5)/(text(i))=text(-)1/2text(i)=1/2text(e)^((3/2pi + k*2pi)text(i))` .

Dit geeft `z=1/2sqrt(2)text(e)^((3/4pi + k*pi)text(i))` .

Dus:
`z_1=0,5-0,5text(i)`
`z_2=text(-)0,5+0,5text(i)`

Opgave 9
a

`(z-text(i))^4=1=text(e)^(k*2pi text(i))` geeft `z=text(e)^(k*1/2pi text(i)) + text(i)`

De oplossingen zijn:
`z_1=text(e)^(0)+text(i)=1+text(i)`
`z_2=text(e)^(1/2pi text(i))+text(i)=text(i)+text(i)=2text(i)`
`z_3=text(e)^(pi text(i))+text(i)=text(-)1+text(i)`
`z_4=text(e)^(1 1/2pi text(i))+text(i)=text(-i)+text(i)=0`

b

Bijvoorbeeld met de abc-formule: z 1 = - 1 6 + 1 6 i 35 ; z 2 = - 1 6 - 1 6 i 35

c

`z^2=8+6text(i)=10text(e)^((0,64...+k*2pi)text(i))` zodat `z=sqrt(10)text(e)^((0,32...+k*pi)text(i))` .
De oplossingen zijn:
`z_1=sqrt(10)text(e)^(0,32...text(i))=3+text(i)`
`z_2=sqrt(13)text(e)^((0,32...+pi)text(i))=text(-)3-text(i)`

d

`(z + 1 - 2text(i))^3 = text(-)2sqrt(3)+2text(i) = 4*text(e)^(text(i)*(5/6 pi + k*2pi))` geeft: `z + 1 - 2text(i) = root[3](4)*text(e)^(text(i)*(5/18 pi + k*2/3 pi))` .

Oplossingen: `z_1~~0,02+3,32text(i)` ; `z_2~~text(-)2,56+2,27text(i)` ; `z_3~~text(-)0,46+0,51text(i)`

e

Op `0` herleiden en de abc-formule toepassen geeft:
`z=(text(-)1-1/2sqrt(2))text(i) vv z=(text(-)1+1/2sqrt(2))text(i)`

f

`z^8+15z^4-16=(z^4+16)(z^4-1)=0` geeft `z^4=text(-)16 vv z^4=1` .

`z^4=16` geeft `z=sqrt(2) + sqrt(2)text(i) vv z=sqrt(2) - sqrt(2)text(i) vv z=text(-)sqrt(2) + sqrt(2)text(i) vv z=text(-)sqrt(2) - sqrt(2)text(i)` .
`z^4=1` geeft `z= 1 vv z=text(i) vv z=text(-)1 vv z=text(-)text(i)` .

Opgave 10
a

`3z+2text(i)`

`=`

`4text(i)z`

`z(3-4text(i))`

`=`

`text(-)2text(i)`

`z`

`=`

`(text(-)2text(i))/(3-4text(i))`

`z`

`=`

`(text(-2)text(i))/(3-4text(i))*(3+4text(i))/(3+4text(i))`

`z`

`=`

`(8-6text(i))/25`

`z`

`=`

`0,32-0,24text(i)`

b

`(z-1)^2=2text(i)=2text(e)^((1/2pi+k*2pi) text(i))` geeft `z-1=sqrt(2)text(e)^((1/4pi+k*pi) text(i))` .

De oplossingen zijn:
`z_1-1=sqrt(2)text(e)^(1/4pi i)=sqrt(2)(1/2sqrt(2)+1/2sqrt(2)i)=2+i`
`z_2-1=sqrt(2)text(e)^(1 1/4pi i)=sqrt(2)(text(-)1/2sqrt(2)-1/2sqrt(2)i)=text(-)i`

c

`z^6=text(-)27=27text(e)^((pi+k*2pi) text(i))` geeft `z=sqrt(3)text(e)^((1/6 pi+k*2/3 pi) text(i))` .
De oplossingen zijn:
`z_1=sqrt(3)text(e)^(1/6pi text(i))=sqrt(3)(1/2sqrt(3)+1/2text(i))=1 1/2+1/2sqrt(3)text(i)`
`z_2=sqrt(3)text(e)^(1/2pi text(i))=sqrt(3)text(i)`
`z_3=sqrt(3)text(e)^(5/6pi text(i))=sqrt(3)(text(-)1/2sqrt(3)+1/2text(i))=text(-)1 1/2+1/2sqrt(3)text(i)`
`z_4=sqrt(3)text(e)^(7/6pi text(i))=sqrt(3)(text(-)1/2sqrt(3)-1/2text(i))=text(-)1 1/2-1/2sqrt(3)text(i)`
`z_5=sqrt(3)text(e)^(9/6pi text(i))=sqrt(3)(text(-)text(i))=text(-)sqrt(3)text(i)`
`z_6=sqrt(3)text(e)^(11/6pi text(i))=sqrt(3)(1/2sqrt(3)-1/2text(i))=1 1/2-1/2sqrt(3)text(i)`

d

`(z+4text(i))(z-4text(i))`

`=`

`16`

`z^2+16`

`=`

`16`

`z^2`

`=`

`0`

`z`

`=`

`0`

Opgave 11

`z^2=text(i)^(text(i))=(text(e)^(1/2pi text(i)))^(text(i))=text(e)^(1/2pi text(i)*text(i))=text(e)^(text(-)1/2pi)` (Merk op dat dit een reëel getal is.)
`z=text(-)sqrt(text(e)^(text(-)1/2pi))~~text(-)0,46 vv z=sqrt(text(e)^(text(-)1/2pi))~~0,46`

Opgave 12De formule van Cardano
De formule van Cardano
a

`t-v=q`
`v=t-q`
Substitueer dit in `t*v=1/27p^3` :
`t*(t-q)=1/27p^3`
`t^2-tq-1/27p^3=0`
Een oplossing is `t=(q+sqrt(q^2+4/27p^3))/2` .
Dit kun je herleiden naar `t=1/2q+sqrt(1/4q^2+1/27p^3)` .
Op dezelfde manier vind je `v=text(-)1/2q+sqrt(1/4q^2+1/27p^3)` , alleen nu substitueer je `t=q+v` in `t*v=1/27p^3` .
Aangezien `x=root(3)(t)-root(3)(v)` , volgt de formule van Cardano.

Bij de andere oplossing krijg je dat `t < 0` en dat kan in deze situatie niet omdat `t` de inhoud van een kubus is. Los van dat kun je deze oplossing wel gebruiken, alleen dan moet je voor `v` ook de andere oplossing nemen aangezien `t-v=q` . Je krijgt dan dezelfde formule.

b

`p=6` en `q=20`
Vul dit in de formule `x = root[3](1/2 q + sqrt(1/4q^2+1/27p^3)) - root[3](text(-)1/2 q + sqrt(1/4q^2 + 1/27p^3))` in.
Dit geeft `x=2` .

c

`x^3+6 x-20 `

`=`

`(x-2 )(x^2+2 x+10 )=0`

`x`

`=`

`2 vv x=text(-)1-3 text(i) vv x=text(-)1+3text(i)`

d

Deel eerst links en rechts van het isgelijkteken door `a` :
`x^3 + b/a x^2 + c/a x + d/a = 0`
Schrijf deze vergelijking door substitutie in de vorm `y^3+py=q` , dan kun je de formule van Cardano gebruiken.
Substitueer daarvoor `x=y-n` in de vergelijking.

`(y-n)^3+b/a(y-n)^2+c/a(y-n)+d/a`

`=`

`0`

`y^3+(text(-)3n+b/a)y^2+(3n-2b/an+c/a)y-n^3+b/an^2-c/an+d/a`

`=`

`0`

Kies `n` zo dat de coëfficiënt voor `y^2` nul wordt:

`(text(-)3n+b/a)y^2`

`=`

`0`

`n`

`=`

`b/(3a)`

Als je `n=b/(3a)` kiest, dan heb je een derdegraadsvergelijking zonder kwadratische term en deze kun je oplossen met behulp van de formule van Cardano. Als je de oplossingen hebt gevonden, trek je daarvan `n` af en je hebt de oplossingen van de oorspronkelijke vergelijking.

e

`x=1/3 vv x=text(-)2 -2sqrt(2)i vv x=text(-)2+2sqrt(2)i`

Opgave 13
a

z = 0 `vv` z = ± 0,5 3 + 1,5 i

b

z = 0,4 + 0,2 i `vv` z = - 0,4 - 0,2 i

c

z 1 1,93 + 0,52 i ; z 2 0,52 + 1,93 i ; z 3 - 1,41 + 1,41 i ; z 4 - 1,93 - 0,52 i ; z 5 - 0,52 - 1,93 i ; z 6 1,41 - 1,41 i

d

z 1 - 0,62 - 3,12 i ; z 2 0,12 + 0,62 i

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