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| 1. | a. | z3
= 8i = 8(cos(1/2π
+ k2π) + isin(1/2π
+ k2π)) z = 2(cos(1/6π + k2/3π) + isin(1/6π + k2/3π)) z1 = 2(cos1/6π + isin1/6π) = √3 + i z2 = 2(cos5/6π + isin5/6π) = -√3 + i z3 = 2(cos11/2π + isin11/2π) = -2i |
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| b. | z2 = 2 + 6i
= √(40)(cos(α
+ k2π) + isin(α
+ k2π)) met tanα
= 6/2 = 3 dus
α = 1,25 z = 401/4 (cos(0,62 + kπ) + isin(0,62 + kπ)) z1 = 2,51 (cos0,62 + isin0,62) = 2,04 + 1,47i z2 = 2,51 (cos3,77 + isin3,77) = -2,04 - 1,47i |
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| c. | z5 = -32
= 32(cos(π + k2π)
+ isin(π + k2π)) z = 2(cos(1/5π + k2/5π) + isin(1/5π + k2/5π)) z1 = 2(cos1/5π + isin1/5π) = 1,62 + 1,18i z2 = 2(cos3/5π + isin3/5π) = -0,62 + 1,90i z3 = 2(cosπ + isinπ) = -2 z4 = 2(cos7/5π + isin7/5π) = -0,62 - 1,90i z5 = 2(cos9/5π + isin9/5π) = 1,62 - 1,18i |
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| d. | z4 = 1 + i =
√2 (cos(1/4π
+ k2π ) + isin(1/4π
+ k2π )) z = 21/8 (cos(1/16π + k 1/2π ) + isin(1/16π + k1/2π )) z1 = 21/8(cos1/16π + isin1/16π ) = 1,07 + 0,21i z2 = 21/8(cos9/16π + isin9/16π ) = -0,21 + 1,07i z3 = 21/8(cos17/16π + isin17/16π ) = -1,07 - 0,21i z4 = 21/8(cos25/16π + isin25/16π ) = 0,21 - 1,07i |
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| e. | z3 = 6 + 4i
= √52(cos(α
+ k2π) + isin(α
+ k2π)) met tanα
= 4/6
dus
α = 0,59 z = 521/6(cos(0,20 + k2/3π) + isin(0,20 + k2/3π)) z1 = 521/6(cos0,20 + isin0,20 ) = 1,89 + 0,38i z2 = 521/6(cos2,29 + i sin2,29) = -1,27 + 1,45i z3 = 521/6(cos4,38 + isin4,38) = -0,62 - 1,83i |
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| f. | (z - i)3 = 3i
+ 5 = √34(cos(α
+ k2π) + isin(α
+ k2π)) met tanα
= 3/5
dus
α = 0,54 z - i = 341/6 (cos(0,18 + k2/3π) + isin(0,18 + k2/3π)) z1 - i = 341/6(cos0,18 + isin0,18) = 1,77 + 0,32i dus z1 = 1,77 + 1,32i z2 - i = 341/6(cos2,27 + isin2,27) = -1,16 + 1,37i dus z2 = -1,16 + 2,37i z3 - i = 341/6(cos4,37 + isin4,37) = -0,61 - 1,69i dus z3 = -0,61 - 0,69i |
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| g. | (2i + 3 + 2z)3 = 2 - 2i =
√8(cos(7/4π
+ k2π) + isin(7/4π
+ k2π)) 2i + 3 + 2z = √2 (cos(7/12π + k2/3π) + isin(7/12π + k2/3π)) 2i + 3 + 2z1 = √2(cos7/12π + isin7/12π) = -0,37 + 1,37i dus 2z1 = -3,37 - 0,63i dus z1 = -1,69 - 0,32i 2i + 3 + 2z2 = √2(cos15/12π + isin15/12π) = -1 - i dus 2z2 = -4 - 3i dus z2 = -2 - 1,5i 2i + 3 + 2z3 = √2(cos23/12π + isin23/12π) = 1,37 - 0,37i dus 2z3 = -1,63 - 2,37i dus z3 = -0,82 - 1,18i |
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| h. | (iz)5 = 2i
= 2(cos(1/2π
+ k2π) + isin(1/2π
+ k2π)) iz = 21/5(cos(1/10π + k2/5π) + isin(1/10π + k2/5π)) iz1 = 21/5(cos1/10π + isin1/10π) = 1,09 + 0,35i dus z1 = 0,35 - 1,09i iz2 = 21/5(cos5/10π + isin5/10π) = 0 + 1,15i dus z2 = 1,15 iz3 = 21/5(cos9/10π + isin9/10π) = -1,09 + 0,35i dus z3 = 0,35 + 1,09i iz4 = 21/5(cos13/10π + isin13/10π) = -0,68 - 0,93i dus z4 = -0,93 + 0,68i iz5 = 21/5(cos17/10π + isin17/10π) = 0,68 - 0,93i dus z5 = -0,93 - 0,68i |
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| 2. | a. | z2
= 2 + 3i = √13(cos(α
+ k2π) + isin(α
+ k2π)) met tanα
= 3/2
dus
α = 0,98 z = 131/4 (cos(0,49 + kπ) + isin(0,49 + kπ)) z1 = 131/4(cos0,49 + isin0,49) = 1,67 + 0,90i z2 = 131/4(cos3,63 + isin3,63) = -1,67 - 0,90i |
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| b. | z2
= 2 + 2i = √8(cos(1/4π
+ k2π) + isin(1/4π
+ k2π)) z = 81/4(cos(1/8π + kπ) + isin(1/8π + kπ) z1 = 81/4(cos1/8π + isin1/8π) = 1,55 + 0,64i z2 = 81/4(cos9/8π + isin9/8π) = -1,55 - 0,64i z2 = 3i = 3(cos(1/2π + k2π) + isin(1/2π + k2π)) z = √3(cos(1/4π + kπ) + isin(1/4π + kπ) z3 = √3(cos1/4π + isin1/4π) = 1,22 + 1,22i z4 = √3(cos5/4π + isin5/4π) = -1,22 - 1,22i Dat geeft samen voor z de mogelijkheden: z = 1,55 + 0,64i + 1,22 + 1,22i = 2,77 + 1,86i z = 1,55 + 0,64i - 1,22 - 1,22i = 0,33 - 0,58i z = -1,55 - 0,64i + 1,22 + 1,22i = -0,33 + 0,58i z = -1,55 - 0,64i - 1,22 - 1,22i - -2,77 - 1,86i |
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| c. | z5
= 1/2 - 2i = √4,25(cos(α
+ 2kπ) + isin(α
+ k2π)) met tanα
= -4 dus
α = -1,32 z = 4,251/10 (cos(-1,32 + k2/5π) + isin(-1,32 + k2/5π)) z1 = 4,251/10(cos-1,32 + isin-1,32) = 0,28 - 1,12i z2 = 4,251/10(cos-0,07 + isin-0,07) = 1,15 - 0,08i z3 = 4,251/10(cos1,19 + isin1,19) = 0,43 + 1,07i z4 = 4,251/10(cos2,44 + isin2,44) = -0,89 + 0,74i z5 = 4,251/10 (cos3,70 + isin3,70) = -0,98 - 0,61i |
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| d. | z3
= (1 + 3i)2 = 1 + 6i - 9 =
-8 + 6i = 10(cos(α + k2π)
+ isin(α + k2π))
met tanα = 6/-8
dus
α = -0,64 z = 101/3(cos(-0,21 + k2/3π) + isin(-0,21 + k2/3π)) z1 = 101/3(cos-0,21 + isin-0,21) = 2,11 - 0,46i z2 = 101/3(cos1,88 + isin1,88) = -0,66 + 2,05i z3 = 101/3(cos3,97 + isin3,97) = -1,45 - 1,59i |
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| 3. | (z2
+ 2i)3 = 4 + 5i = √41(cos(α + k2π)
+ isin(α + k2π)) met
tanα = 5/4 dus
α = 0,896 z2 + 2i = 411/6(cos(0,30 + k2/3π) + isin(0,30 + k2/3π)) z12 + 2i = 411/6(cos0,30 + isin0,30) = 1,775 + 0,546i dus z12 = 1,775 - 1,455i z22 + 2i = 411/6(cos2,39 + isin2,39) = -1,361 + 1,264i dus z22 = -1,361 - 0,736i z32 + 2i = 411/6(cos4,49 + isin4,49) = -0,414 - 1,810i dus z32 = -0,414 - 3,810i Dat zijn drie nieuwe opgaven: |
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| z12
= 1,775 - 1,455i = 2,295(cos(α + k2π)
+ isin(α + k2π))
met tanα = -1,455/1,775
dus
α = -0,687 z1 = 2,2950,5(cos(-0,343 + kπ) + isin(-0,343 + kπ)) Dat geeft z = 2,2950,5(cos-0,343 + isin-0,343) = 1,43 - 0,51i z = 2,2950,5(cos2,798 + isin2,798) = -1,43 + 0,51i z22 = -1,361 - 0,736i = 1,547(cos(α + k2π) + isin(α + k2π)) met tanα = -0,736/-1,361 dus α = -2,646 (in IV) z2 = 1,5470,5(cos(-1,323 + kπ) + isin(-1,323 + kπ) Dat geeft z = 1,5470,5(cos-1,323 + isin-1,323) = 0,31 - 1,21i z = 1,5470,5(cos1,819 + isin1,819) = -0,31 + 1,21i z32 = -0,414 - 3,810i = 3,832(cos(α + k2π) + isin(α + k2π)) met tanα = -3,810/-0,414 dus α = 4,604 (in IV) z3 = 3,8320,5(cos(2,302 + kπ) + isin(2,302 + kπ)) Dat geeft: z = 3,8320,5(cos2,302 + isin2,302) = -1,31 + 1,46i z = 3,8320,5(cos5,444 + isin5,444) = 1,31 - 1,46i De zes rode getallen zijn oplossingen voor z |
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| 4. | a. | 1/2√3
+ 1/2i
heeft r = 1 en hoek
φ
= 1/6π De wortel daarvan (tenminste de ene) heeft dan r = 1 en hoek 1/12π dus dat is het getal cos(1/12π) + isin(1/12π) |
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| b. | (a + bi)2
= 1/2√3
+ 1/2i
a2 + 2abi - b2 = 1/2√3 + 1/2i De reλle delen van beide kanten zijn gelijk: a2 - b2 = 1/2√3 De imaginaire delen van beide kanten zijn gelijk: 2ab = 1/2 |
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| c. | Uit de tweede
vergelijking volgt b = 1/4a en dat kun je
invullen in de eerste: a2 - 1/(16a2)= 1/2√3 Noem a2 = p dan staat hier p - 1/(16p) = 1/2√3 16p2 - 1 = 8p√3 16p2 - 8p√3 - 1 = 0 ABC-formule: p = (8√3 ± √(192 + 64))/32 = (8√3 ± 16)/32 = 1/4√3 ± 1/2 Maar omdat p positief moet zijn (het is immers een kwadraat) voldoet alleen p = 1/2 + 1/4√3 Dus a2 = 1/2 + 1/4√3 |
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| d. | cos1/12π = a = √(1/2 + 1/4√3) | ||||
| 5. | a. | Elke keer als je met z vermenigvuldigt wordt de afstand tot de oorsprong (r) groter, want de spiraal draait weg van O. Dus is rZ > 1 | |||
| b. | z8
heeft afstand tot de oorsprong 2, dus rz8
= 2 dus rZ = 21/8 z10 heeft hoek φ = 11/2π, dus z heeft hoek 1,5p/10 = 0,15π Beiden geldt als z = 21/8 (cos0,15π + isin0,15π) |
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© h.hofstede (h.hofstede@hogeland.nl) |
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