So the complex number z is given as
$z=1-cos(2\theta)-isin(2\theta)$
And the question is to find the argument of z in terms of $\theta$.
So, I used the formula for calculating the argument of a complex number:
$arg(z)=arctan (\frac{sin(2\theta)}{1-cos(2\theta)})$
$=arctan (\frac{2sin\theta cos\theta}{1-(1-2sin^2\theta)})$
$=arctan (\frac{2sin\theta cos\theta}{2sin^2\theta})$
$=arctan (\frac{cos\theta}{sin\theta})$
$=arctan (cot\theta)$
$=arctan (tan(\frac{\pi}{2}-\theta))$
$=\frac{\pi}{2}-\theta$
But apparently the answer is $\theta-\frac{\pi}{2}$. Where might have I gone wrong?
Edit: the domain given in the question for $\theta$ was $0≤\theta≤\pi$
Best Answer
\begin{align*} 1-\cos2\theta -i\sin2\theta&=2\sin^2\theta-2i\sin\theta \cos\theta\\ &=2\sin\theta(\sin\theta-i\cos\theta)\\ &=2\sin\theta\left[\cos\left(\theta-\frac{\pi}{2}\right)+i\sin\left(\theta-\frac{\pi}{2}\right)\right]\\ &=-2\sin\theta\left[\cos\left(\theta+\frac{\pi}{2}\right)+i\sin\left(\theta+\frac{\pi}{2}\right)\right] \end{align*}
If $\sin\theta>0$, the argument is $\displaystyle \theta-\frac{\pi}{2}$.
If $\sin\theta<0$, the argument is $\displaystyle \theta+\frac{\pi}{2}$.