# Finding the argument of $z=1-\cos(2\theta)-i\sin(2\theta)$

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 $0≤\theta≤\pi$

• Any domain given for $\theta$? Apr 26, 2018 at 9:38
• You missed a minus sign in the first step. Apr 26, 2018 at 9:41

If $\sin\theta>0$, the argument is $\displaystyle \theta-\frac{\pi}{2}$.
If $\sin\theta<0$, the argument is $\displaystyle \theta+\frac{\pi}{2}$.
For the first step, the $$\sin2θ$$ should be changed into $$-\sin2θ$$, then it would be $$\arctan(-\cotθ)$$, and the final answer would be $$θ−\fracπ2$$ as well.