I have a parabola $y^2=4x$, and I have to rotate this by $120^\circ$ in the anticlockwise direction. What my book does is, it takes a point $(x,y)$ in the complex form, ie, $z=x+\iota y$, and then by rotation, the new point is $z_1=ze^{\frac{2\pi}{3}} = \bigg(-\frac{x}{2}-\frac{\sqrt3 y}{2}\bigg)+\iota \bigg(\frac{\sqrt3x}{2}-\frac{y}{2}\bigg)$
So the equation of the new curve, the book says is
$\bigg(\frac{\sqrt3x}{2}-\frac{y}{2}\bigg)^2=4\bigg(-\frac{x}{2}-\frac{\sqrt3 y}{2}\bigg)$
But the axis of this curve is $\sqrt3 x = y$, which has a slope of $\sqrt3$, ie it makes an angle of $60^\circ$ with the positive direction of the $x$-axis.
My question is, is rotating the point on the curve by an angle (as the book does), same as rotating the axis of a curve by that angle? If not, why? In what "sense" is this rotation?
I believe they may have mixed up the original and new co-ordinates in the book.
Just as, if you want to translate the graph of $y=f(x)$ by $a\in\mathbb R$ horizontally, the resulting graph has the equation $y=f(x-a)$ and not $y=f(x+a)$.
The correct solution would be not to express $z_1=x_1+iy_1=ze^{2\pi i/3}$, but to express $z=x+iy=z_1e^{-2\pi i/3}$, which gives you $x, y$ expressed via $x_1, y_1$, and then, the original equation $y^2=4x$, by substituting, gives you the equation of the rotated curve in terms of $x_1, y_1$.