complex analysis question－百科全書

標題:

complex analysis question

發問:

Let A={z∈C:1/2
最佳解答:

(1) f(z) = z + 1/z Let z = x + y i For circle of radius r, x^2 + y^2 = r^2 f(z) = x + yi + 1/(x+yi) = (x + yi) + (x – yi)/(x^2 + y^2) = (x+ yi) + (x – yi)/r^2 = (1 + 1/r^2)x + (1 – 1/r^2)y i = X + Yi When r = 1, X + Yi = 2x the image is the line segment from X = -2 to 2 and Y = 0 Where X = (1 + 1/r^2)x or x = X/(1 + 1/r^2) And Y = (1 – 1/r^2)y or y = Y/(1 – 1/r^2) x^2 + y^2 = X^2/(1 + 1/r^2)^2 + Y^2/(1 – 1/r^2)^2 r^2 = X^2/(1 + 1/r^2)^2 + Y^2/(1 – 1/r^2)^2 X^2/(r + 1/r)^2 + Y^2/(r – 1/r)^2 = 1 which is an ellipse (2) when r = 1/2, the image is X^2/(1/2 + 2)^2 + Y^2(1/2 – 2)^2 = 1 or X^2/2.5^2 + Y^2/1.5^2 = 1 When r = 2, the image is X^2/(2 + 1/2)^2 + Y^2/(2 – 1/2)^2 = 1 X^2/2.5^2 + Y^2/1.5^2 = 1 the same ellipse as for the case r = 1/2 Consider H = h^2 = (r + 1/r)^2 = r^2 + 2 + 1/r^2 dH/dr = 2r – 2/r^3 d^H/dr^2 = 2 + 6/r^4 When dH/dr = 0, r = 1, and d^2H/dr^2 = 8 > 0 There h^2 only has min at r = 1 Similarly consider K = k^2 = (r – 1/r)^2, it can be shown that k^2 only has minimum when r = 1. That is to say the image ellipse of any circle with radius between ? and 2 always has a dimension less than or equal to the cases when the radius ? or 2. Hence f(A) is the interior of the ellipse X^2/2.5^2 + Y^2/1.5^2 = 1

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