If $|z|=1$ and $\arg(z)=\theta$, $0<\theta<\frac{\pi}{2}$, find $\arg(\frac{2}{1-z^2})$. $\newcommand{\cis}{\operatorname{cis}}$If $|z|=1$ and $\arg(z)=\theta$, $0<\theta<\frac{\pi}{2}$, find $\arg(\frac{2}{1-z^2})$.
First, because $|z|=1$, then I can express $z$ as $z=\cis(\theta)$. Then I need to get rid of the one in the numerator: $1-z^2=1-(\cis(\theta))^2=1-\cis(2\theta)=[\cdots]=-2i\sin(\theta)\cis(\theta)$ (after some transformation).
Then $\displaystyle \frac{2}{1-z^2}=\frac{2}{-2i\sin(\theta)\cis(\theta)}=\frac{i}{\sin(\theta)(\cis(\theta)}=\frac{1}{\sin(\theta)}\cdot\frac{\cis(\frac{\pi}{2})}{\cis(\pi)}=\frac{1}{\sin(\theta)}\cdot\cis\left(\frac{\pi}{2}-\theta\right)$.
So the argument should be $\frac{\pi}{2}-\theta$. However, my textbook gives the answer $\theta-\frac{\pi}{2}$. It also says that the mod is $\csc(\theta)$, which agrees with $\frac{1}{\sin(\theta)}$, but the argument is different - what am I doing wrong?
 A: It amounts to finding the argument of $\displaystyle\frac1{1-z^2}$. Now, if $\def\I{\mathrm i\mkern 1mu}\def\E{\mathrm e}z=\E^{\I\theta}$, then
$$\frac1{1-z^2} =\frac1{1-\mathrm e^{2\I\theta}}=\frac{-\E^{\I\theta}}{\E^{\I\theta}-\E^{-\I\theta}}=\frac2{\sin\theta}(-\I\E^{\I\theta})=\frac2{\sin\theta}\E^{\I(\theta-\tfrac\pi2)}.$$
Hence
$$\arg\frac2{1-z^2}\equiv\theta-\frac\pi2\mod2\pi.$$
A: $$1-z^2=1-(\cos2\theta+i\sin2\theta)=2\sin^2\theta+i2\sin\theta\cos\theta=2i\sin\theta(\cos\theta-i\sin\theta)$$
$$\dfrac2{1-z^2}=\dfrac2{2i\sin\theta(\cos\theta-i\sin\theta)}=\dfrac{-i(\cos\theta+i\sin\theta)}{\sin\theta}=1-i\cot\theta$$
If $\tan A=\dfrac{-\cot\theta}1=\tan\left(\dfrac\pi2+\theta\right)$
$A=n\pi+\dfrac\pi2+\theta$ where $n$ is any integer
As $0<\theta<\dfrac\pi2,n\pi+\dfrac\pi2<A<n\pi+\dfrac\pi2+\dfrac\pi2$
If $A\in\left(-\dfrac\pi2,\dfrac\pi2\right); n=-1\implies A=(-1)\pi+\dfrac\pi2+\theta=\theta-\dfrac\pi2$
Using this, 
 arg$\left(\dfrac2{1-z^2}\right)=\theta-\dfrac\pi2$
A: So $\;z\in S^1\;$ and on the arc in the first quadrant, thus $\;z=\cos\theta+i\sin\theta\;\implies\;$ 
$$\frac2{1-z^2}=\frac2{1-\cos2\theta-i\sin2\theta}=\frac{2-2\cos2\theta+2i\sin2\theta}{2(1-\cos2\theta)}=$$
$$=1+\frac{\sin2\theta}{1-\cos2\theta}i\implies\arg\left(\frac2{1-z^2}\right)=\arctan\frac{\cos\theta}{\sin\theta}=$$
$$=\frac\pi2-\arctan\frac{\sin\theta}{\cos\theta}=\frac\pi2-\theta$$
and I can't see how the textbook gives the additive inverse of this since the argument of the arctangent is positive, so I'd say you're right and the textbook is wrong.
