Bonjour,
z = x + iy
z' = x' + iy'
z' = (1 + i√3)z
⇔ z' = (1 + i√3)(x + iy)
⇔ z' = (x - √3y) + (√3x + y)i
⇒ x' = x - √3y
y' = √3x + y
⇒ x = x' + √3y
y = y' - √3(x' + √3y)
⇒ 4y = -√3x' + y'
y = -√3x'/4 + y'/4
⇒ x = x' + √3(-√3x'/4 + y'/4)
x = (1 - 3/4)x' + √3y'/4
x = x'/4 + √3y'/4
x² - 3y² = 3
⇔ 1/16(x'² + 2√3x'y' + 3y'²) - 3/16(3x'² -2√3x'y' + y'²) = 3
⇔ (1 - 9)x'² + (2√3 + 6√3)x'y' + (3 - 3)y'² = 48
⇔ -8x'² + 8√3x'y' = 48
⇔ x'² + √3x'y' = 6
⇔ y' = (6 - x'²)/√3x'
⇔ y' = -√3/x' + 2√3x' hyperbole
