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| 1. | a. | sin(1/6π
+ i) = 1/2i(eπi/6 - 1 + e-πi/6 + 1) = 1/2i • (e-1 • (cos(π/6) + isin(π/6)) + e(cos(-π/6) + isin(-π/6))) = 1/2i • (e-1 • (1/2√3 + 1/2i) + e • (1/2√3 - 1/2i)) = e-1 • (-1/4i√3 + 1/4) + e • (-1/4i√3 - 1/4) = -0,59 - 1,34i |
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| b. | cos(π
+ 4i) = 1/2(eiπ - 4 + e-iπ + 4) = 1/2(e-4 • eiπ + e4 • e-iπ) = 1/2(e-4 • -1 + e4 • -1) = 27,81 |
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| c. | sin(3i) = 1/2i • ( e-3 - e3) = 1/2i • -20,03 = 10,02i |
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| d. | cos(2i - 1/2π) = 1/2(e-2 - 0,5πi + e2 + 0,5πi) = 1/2(e-2 • e-0,5πi ) + e2 • e0,5πi = 1/2(e-2 • -i + e2 • i) = 3,63i |
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| e. | cos(-2i) = 1/2(e2 + e-2) = 3,76 |
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| f. | sin(1 + 3i) = 1/2i • (ei - 3 + e-i + 3 ) = 1/2i • (e-3 • (cos1 + isin1) + e3 • (cos(-1) + isin(-1)) = 1/2i • (0,027 + 0,042i + 10,852 - 16,901i) = 1/2i • (10,879 - 16,859i) = -8,43 - 5,44i |
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| 2. |
 |
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| =
1/4(4eiz
• eiz ) = 1/4 • 4 • 1 = 1 |
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| 3. | a. | 1/2(eiz
+ e-iz ) = 2
vermenigvuldig beide kanten met 2eiz e2iz + 1 = 4eiz (eiz)2 - 4eiz + 1 = 0 Noem eiz = p dan staat er p2 - 4p + 1 = 0 ABC-formule: p = (4 ± √(16 - 4))/2 = (4 ± √12)/2 = (4 ± 2√3)/2 = 2 ± √3 Dus eiz = 2 ± √3 |
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| b. | r • eiφ =
eiz ln(r • eiφ) = ln(eiz) lnr + lneiφ = iz lnr + iφ = iz 1/i • lnr + φ = z -ilnr + φ = z |
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| c. | eiz
= 2 +
√3 = (2 +
√3) • e0i
geeft dan z = -iln(2
+ √3) eiz = 2 - √3 = (2 - √3) • eπi geeft dan z = iln(2 - √3) + π |
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© h.hofstede (h.hofstede@hogeland.nl) |
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