Prove that the following set of differential equations yield a circle

Question

So I have the following differential equations:

$$\dot{x} = \omega y \ \ \ \ (i) $$ $$\dot{y}= -\omega x \ \ \ (ii)$$

Now, my professor considered them as complex variables and solved in the following way:

$$\dot{x}+i\dot{y} = -\omega (y-ix) \ \ \ \ (iii) $$ $$\implies \dot{z} = -i\omega z \ \ \ \ (iv)$$

Which can be further solved to prove that it gives a circle.

My doubt is regarding the third equation. That is shouldn't the equation come out to be as :

$$ x+iy = -\omega(iy -x )$$

Or, is it that we simply are multiplying i to the equation then adding them ?

Answer 1

Setting $z = x + iy$ we have $$ \dot{z} = \dot{x} + i \dot{y} = \omega y + i (-\omega x) = -i\omega (x + iy) = -i\omega z $$ with solutions $z = z_0 e^{-i\omega t},$ where $z_0 \in \mathbb{C}.$

Then writing $z_0 = R e^{i\phi}$ we get $z = R e^{-i(\omega t-\phi)}$ so $$\begin{cases} x = \operatorname{Re} z = R \cos(\omega t - \phi) \\ y = \operatorname{Im} z = R \sin(\omega t - \phi) \end{cases}$$ which is a parameterization of a circle with center $(0,0)$ and radius $R$.