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標題:
complex number - 0定π
發問:
z^3=1 tanθ = 0 z^4= -1 tanθ = π 係唔係ar? 係咪次次z^n=1就係tanθ = 0 z^n= -1就係tanθ = π 關唔關n事....姐係odd/even果d
最佳解答:
z^3 = 1 rcis3θ = 1 r = 1 cos3θ= 1 = cos 0 sin3θ= 0 = sin 0 3θ = 2nπ θ = 2nπ/3 where n is integer (tanθ= 0 or -√3 or √3) z^4 = -1 rcis4θ = -1 r = 1 cos4θ= -1 = cos π sin4θ= 0 = sin π 4θ = (2n+1)π θ = (2n+1)π/4 where n is integer (tanθ= -1 or 1 ) { Why do you think that tanθ= 0 or π? }
其他解答:
complex number - 0定π
發問:
z^3=1 tanθ = 0 z^4= -1 tanθ = π 係唔係ar? 係咪次次z^n=1就係tanθ = 0 z^n= -1就係tanθ = π 關唔關n事....姐係odd/even果d
最佳解答:
z^3 = 1 rcis3θ = 1 r = 1 cos3θ= 1 = cos 0 sin3θ= 0 = sin 0 3θ = 2nπ θ = 2nπ/3 where n is integer (tanθ= 0 or -√3 or √3) z^4 = -1 rcis4θ = -1 r = 1 cos4θ= -1 = cos π sin4θ= 0 = sin π 4θ = (2n+1)π θ = (2n+1)π/4 where n is integer (tanθ= -1 or 1 ) { Why do you think that tanθ= 0 or π? }
其他解答:
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