Is there a way to recover $z = -2i\sin\theta$ where $0 < \theta < \pi$ into the form $r(\cos\theta+i\sin\theta)$?

Question

Is there a way to recover $z = -2i\sin\theta$ where $0 < \theta < \pi$ into the form $r(\cos\theta+i\sin\theta)$?

Because it is confusing as $r$ is not $2$ here and should be $r = 2\sin\theta$.

The argument of $-2i\sin\theta$ should be $-\frac{1}{2}\pi$ but i just cannot seem to put it in polar form because $\theta >0$... Am I messing up with the variables here?

Answer 1

If $-2i\sin\theta=r(\cos t+i\sin t),r\ge0$

Clearly $r\ne0,\cos t=0,\sin t =-1,r=2$ as $r>0$

$\implies t=2m\pi -\dfrac\pi2$ where $m$ is any integer

Answer 2

$$z=-2i\sin(\theta)=2\sin(\theta)e^{-\frac{\pi} 2}$$ Because $0<\theta<\pi$, we know that $\sin\theta>0$

So the polar form is $$z=r(\cos \alpha +i\sin \alpha)$$ With $r=2\sin(\theta)$ and $\alpha=-\frac \pi 2$.