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

## 2 Answers

\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 $\displaystyle \theta+\frac{\pi}{2}$.

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.