If $z=r\cos\theta +ir\sin \theta$, show that $\dfrac{dz}{dt}=r\dfrac{d\theta}{dt}(-\sin\theta+i\cos \theta)$ and that if $\dfrac{d\theta}{dt}$ is constant, then $\arg\dfrac{dz}{dt}=\theta-\pi/2$.
I managed to do everything apart from the argument, which I calculated to be $\theta+\pi/2$ instead. Can anyone help?
The answer says arg(-sinx+icosx)=arctan(cosx/-sinx)=arctan(-cotx)=-arctan(cotx)=-arctan(tan(pi/2-x))=-(pi/2-x)=x-pi/2. What is wrong with this approach?
$z=r\cos(\theta) + ir\sin(\theta) = re^{i\theta}$
$\frac{dz}{dt}=rie^{i\theta}\frac{d\theta}{dt} = re^{\frac{i\pi}{2}}e^{i\theta}\frac{d\theta}{dt} = r\frac{d\theta}{dt}e^{i(\theta + \frac{\pi}{2})}$
Therefore, arg$\frac{dz}{dt}= \theta + \frac{\pi}{2}$ and you are correct.