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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

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