✔ 最佳答案
(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 ½ and 2 always has a dimension less than or equal to the cases when the radius ½ or 2.
Hence f(A) is the interior of the ellipse X^2/2.5^2 + Y^2/1.5^2 = 1
