香港特色

標題:

complex numbers

發問:

Find all solutions to Z^5 + 32 = 0

最佳解答:

As the degree of z is 5, we expect 5 solutions. By representing complex numbers in an argand graph, the solutions have the seperation of 360 /5 = 72 (degrees). The first solution : z = (-32)^(1/5) = -2 = -2 + 0i = -2 (1+0i) Note that complex number takes the general form z = cos(theta) + sin(theta)i Hence, for z = -2 + 0i, sin(theta) = 0 , theta = 0 (degree) The second solution: z = -2 (cos(0+72) + sin(0+72)) = -2 (cos (72) + isin(72)) Similarly, for third solution, theta = 72 *2 = 144 Third solution: -2 (cos (144) + isin(144)) Fourth solution:-2 (cos (216) + isin(216)) Fifth solution:-2 (cos (288) + isin(288)) where theta is in degree

其他解答:

Let cis a = cos a + i sin a z^5 + 32 = 0 z^5 = -32 = 2^5 (1) = 2^5 cis π z = 2 cis [(2k+1)π/5] , k=0,1,2,3,4 i.e. z = 2 cos (π/5) + i sin (π/5) ; or z = 2 cos (3π/5) + i sin (3π/5) ; or z = 2 cos (5π/5) + i sin (5π/5) = -2 ; or z = 2 cos (7π/5) + i sin (7π/5) ; or z = 2 cos (9π/5) + i sin (9π/5) 2010-05-25 03:30:40 補充： 更正： 第三行 z^5 = -32 = 2^5 (-1)|||||.............. z = -2 .............. 2010-05-25 01:59:27 補充： Z^5 + 32 = 0 Z^5 = -32 Z^5 = -2^5 z = -2

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