Stelle die folgenden komplexen Zahlen in der arithmetischen Form ´a + bi´ (mit ´a, b in RR´) und in der trigonometrischen Form ´r * e^(i alpha)´ (´r in R^+´, Winkel im Bogenmaß, ´0 <= alpha < 2pi´) dar.
- ´(−12 + 18 bb i)/(1 + 5 bb i)´
- ´2e^(bb i (7pi)/6) * (sqrt(3) + bb i)´
- ´(4 + sqrt(48) bb i)^8´
(Zweierpotenzen müssen nicht ausmultipliziert werden.)
"´r \* e^(i alpha) = r(cos(alpha) + i sin(alpha)) = r \* cos(alpha) + r \* sin(alpha)´\n\n´a_1 + a_2 i = sqrt(a_1^2 + a_2^2) * (a_1/(sqrt(a_1^2 + a_2^2)) + a_2/(sqrt(a_1^2 + a_2^2)) i)´ mit ´0 <= alpha <= 2pi´\n\n\nTeil 1\n\n´z = (-12 + 18i)/(1 + 5i) = ((-12 + 18i)(1-5i))/((1 + 5i)(1-5i))´\n´= (-12 + 60i + 18i + 90i)/(1 + 25) = (78 + 78i)/26 = 3 + 3i´\n\n´|z| = sqrt(3^2 + 3^2) = sqrt(18)´\n\n´z = sqrt(18) (3/sqrt(18) + 3/sqrt(18) i)´\n´= sqrt(18) (3/(3 sqrt(2)) + 3/(3 sqrt(2)) i)´\n´= sqrt(18) (1/sqrt(2) + 1/sqrt(2) i)´\n´= sqrt(18) (sqrt(2)/2 + sqrt(2)/2 i)´\n´= sqrt(18) (cos(pi/4) + sin(pi/4) i)´\n´= 3 sqrt(2) * e^(i pi/4)´\n\n\nTeil 2\n\n´z_0 = sqrt(3) + i´\n´|z_0| = sqrt(3 + 1) = 2´\n\n´z_0 = 2 (sqrt(3)/2 + 1/2 i)´\n´= 2 (cos(pi/6) + sin(pi/6) i)´\n´= 2e^(i pi/6)´\n\n´2e^(i (7pi)/6) * 2e^(i pi/6)´\n´= 4e^(i((7pi)/6 + pi/6))´\n´= 4e^(i (8pi)/6)´\n´= 4e^(i (4pi)/3)´\n\n´= 4 (cos((4pi)/3) + sin((4pi)/3) i)´\n´= 4 (- 1/2 - 1/2 i)´\n´= -2 - 2 i´\n\nSolution seems to be wrong!\n\n\nTeil 3\n\n´(4 + sqrt(48)i)^8 = sum_(n=0)^8 ((8),(k)) 4^k (sqrt(48)i)^(8-k)´\n\nZu aufwendig!\n\n´4 + sqrt(48)i = 8(1/2 + (4 sqrt(3))/8 i)´\n´= 8(cos(pi/3) + sin(pi/3) i)´\n´= 8e^(i pi/3)´\n\n´=> 4 + sqrt(48)i = 8e^(i pi/3)^8 = 8^8 e^(i (8pi)/3)´\n´= 8^8 (cos(8/3 pi) + sin(8/3 pi) i)´\n´= 8^8 (-1/2 + sqrt(3)/2 i)´\n´= 2^24 (-1/2 + sqrt(3)/2 i)´\n´= -2^23 + 2^23 sqrt(3) i)´"
- ´3 + 3i´
- ´3 sqrt(2) * e^(i pi/4)´
- ´2e^(i pi/6)´
- ´-2 - 2 i´ Solution seems to be wrong!
- ´8e^(i pi/3)´
- ´-2^23 + 2^23 sqrt(3) i)´
HPI, 2014-07-07, Mathe 2, Aufgabe 54
2014-07-27 23:20:32 UTC
2014-07-27 23:20:32 UTC