Visualizing rotation of axes in coordinate geometry using complex numbers

Question

I am currently studying rotation of axes, and one of the methods in my book states that the rotation can also be remembered by using complex numbers rotation method to make the substitution easier to memorize.

I started by taking two perpendicular axes $x$ and $y$, and let a point on this plane be $x+ iy$. Now, I'm rotating the axes by an angle $\theta$ in the counterclockwise direction. Let the new axes be $y'$ and $x'$, and a point, with respect to the new axes be $x'+iy'$. Now, when I rotate this axes back to the old one (by a clockwise rotation of the same angle $\theta$), I must get back my original point, i.e, $x+iy$.

Using this, I get $$(x'+iy')e^{-i\theta}=x+iy$$

$($Using the negative sign in RHS to denote a clockwise rotation of the new axes to get back ot the old axes$)$ $$\implies x'+iy'=(x+iy)e^{i\theta}$$ $$\implies x'+iy'=(x+iy)(\cos \theta + i \sin \theta)$$ From here, comparing the real and imaginary parts of the equation, I get the following two results

$$x'=x\cos \theta - y\sin \theta$$ $$y'=x\sin \theta + y \cos \theta$$

But, my book states, in order to turn the axes counterclockwise, through an angle theta, we must substitute, $$x=x'\cos \theta - y'\sin \theta$$ $$y=x'\sin \theta + y' \cos \theta$$

I get the equation in similar forms but I think there is a mistake in my analysis which causes me to get the "opposite" answer. Can someone explain the error in my analysis?

Answer 1

This always used to make me crazy too, so it is best to discuss the transformations systematically. Call the point $ \ z \ = \ x + i·y \ = \ r·e^{ \ i·\phi} \ = \ r·(\cos \phi + i·\sin \phi) \ \ . $ So we have $ \ x \ = \ r·\cos \phi \ \ , \ \ y \ = \ r·\sin \phi \ \ . $

In the new coordinate system resulting from rotating the axes counter-clockwise by $ \ \theta \ \ , $ we obtain $$ z \ \ = \ \ x' + i·y' \ \ = \ \ r·e^{ \ i·(\phi - \theta)} \ = \ r·[ \ \cos (\phi - \theta) \ + \ i·\sin (\phi - \theta) \ ] $$ $$ = \ r·[ \ (\cos \phi · \cos \theta \ + \ \sin \phi · \sin \theta) \ + \ i·(\sin \phi · \cos \theta \ - \ \cos \phi · \sin \theta) \ ] $$ $$ \Rightarrow \ \ x' \ \ = \ \ \underbrace{r · \cos \phi}_{x} · \cos \theta \ + \ \underbrace{r · \sin \phi}_{y} · \sin \theta \ \ = \ \ x · \cos \theta \ + \ y · \sin \theta \ \ , $$ $$ y' \ \ = \ \ \underbrace{r · \sin \phi}_{y} · \cos \theta \ - \ \underbrace{r · \cos \phi}_{x} · \sin \theta \ \ = \ \ -x · \sin \theta \ + \ y · \cos \theta \ \ . $$ [Notice, incidentally, that if we wrote the transformation in terms of a matrix, its determinant would be $ \ \cos \theta · \cos \theta - (-\sin \theta) · \sin \theta \ = \ \cos^2 \theta + \sin^2 \theta \ = \ 1 \ \ , $ so we have pure rotation without "re-scaling" or change in the "handedness" of the coordinate system.]

Let us now re-name the new angle as $ \ \psi \ = \ \phi - \theta \ \ $ and make the clockwise rotation by $ \ \theta \ \ $ to produce $$ z \ \ = \ \ x'' + i·y'' \ \ = \ \ r·e^{ \ i·(\psi + \theta)} \ = \ r·[ \ \cos (\psi + \theta) \ + \ i·\sin (\psi + \theta) \ ] \ \ ; $$ we'll "pretend" we don't know how this is supposed to end up and see what this transformation produces: $$ z \ \ = \ r·[ \ (\cos \psi · \cos \theta \ - \ \sin \psi · \sin \theta) \ + \ i·(\sin \psi · \cos \theta \ + \ \cos \psi · \sin \theta) \ ] $$ $$ \Rightarrow \ \ x'' \ \ = \ \ \underbrace{r · \cos \psi}_{x'} · \cos \theta \ - \ \underbrace{r · \sin \psi}_{y'} · \sin \theta \ \ = \ \ x' · \cos \theta \ - \ y' · \sin \theta \ \ , $$ $$ y'' \ \ = \ \ \underbrace{r · \sin \psi}_{y'} · \cos \theta \ + \ \underbrace{r · \cos \psi}_{x'} · \sin \theta \ \ = \ \ x' · \sin \theta \ + \ y' · \cos \theta \ \ , $$ which is the transformation your book presents.

Here again, we see that the determinant of the transformation matrix is $ \ \cos^2 \theta + \sin^2 \theta \ = \ 1 \ \ . $ We can think of this as a counter-clockwise rotation by $ \ (-\theta) \ \ , $ which is $$ x'' \ \ = \ \ x' · \cos (-\theta) \ + \ y · \sin (-\theta) \ \ = \ \ x' · \cos \theta \ + \ y' · (-\sin \theta) \ \ , $$ $$ y'' \ \ = \ \ = \ \ -x' · \sin (-\theta) \ + \ y' · \cos (-\theta) \ \ = \ \ -x' · (-\sin \theta) \ + \ y' · \cos \theta \ \ . $$

If we insert the "primed" coordinates into this transformation, we find $$ x'' \ \ = \ \ [ \ x · \cos \theta \ + \ y · \sin \theta \ ] \ · \cos \theta \ + \ [ \ -x · \sin \theta \ + \ y · \cos \theta \ ] · (-\sin \theta) $$ $$ = \ \ x · ( \ \cos^2 \theta \ + \ \sin^2 \theta \ ) \ + \ y · (\sin \theta · \cos \theta \ - \ \cos \theta · \sin \theta) \ \ = \ \ x \ \ , $$ $$ y'' \ \ = \ \ -[ \ x · \cos \theta \ + \ y · \sin \theta \ ] · (-\sin \theta) \ + \ [ \ -x · \sin \theta \ + \ y · \cos \theta \ ] · \cos \theta $$ $$ = \ \ x · [ \ (-\cos \theta)·(- \sin \theta) \ + \ (- \sin \theta) · \cos \theta \ ] $$ $$ + \ y · [ \ (-\sin \theta) · (-\sin \theta) \ + \ \cos \theta · \cos \theta \ ] \ \ = \ \ y \ \ . $$

So, the CCW and CW rotations do "invert" one another and we get back our original coordinates $ \ z \ = \ x'' + i·y'' \ = \ r·e^{ \ i·(\phi - \theta + \theta)} \ = \ x + i·y \ \ . $

I think you may just have confused yourself (or the book confused you) about what you and what the book are calling the "primed coordinates."

Answer 2

Your approach sends $x=1,\,y=0$ to $x^\prime=\cos\theta,\,y^\prime=\sin\theta$ and $x=0,\,y=1$ to $x^\prime=-\sin\theta,\,y^\prime=\cos\theta$, i.e. rotates the point anticlockwise by $\theta$ while preserving the original choice of coordinate axes, which is an active transformation. The problem you should have solved is computing the original point's coordinates in a new system, which is a passive transformation. Since we want to rotate the axes anticlockwise $\theta$, it looks the same as if the point has rotated anticlockwise $-\theta$, which explains why an overall sign change for $\theta$ turns your solution attempt into the correct one.