Given that $z= \cos \theta + i\sin \theta$, prove that $\Re\left\{\dfrac{z-1}{z+1}\right\}=0$ Given that $z= \cos \theta + i\sin \theta$, prove that $\Re\left\{\dfrac{z-1}{z+1}\right\}=0$
How would I do this?
 A: $$\frac{z-1}{z+1}=\frac{z-1}{z+1}\frac{\overline z+1}{\overline z+1}=\frac{|z|^2-1+2i\,\text{Im}\,z}{|z+1|^2}\implies\text{Re}\,\left(\frac{z-1}{z+1}\right)=\frac{|z|^2-1}{|z+1|^2}$$
But we know that $\;z=\cos\theta+i\sin\theta\implies |z|=1\;$ , so...
A: If $w=a+ib,\bar w=a-ib\implies w+\bar w=2a$
So, $w$ will be purely imaginary iff $ w+\bar w=0$
Here $$\frac{z-1}{z+1}+\overline{\left(\frac{z-1}{z+1}\right)}$$
$$=\frac{z-1}{z+1}+\frac{\bar z-1}{\bar z+1}=\frac{z\bar z-\bar z+z-1+(z\bar z+\bar z-z-1)}{(z+1)(\bar z+1)}$$
$$=\frac{2(z\bar z-1)}{z\bar z+(z+\bar z)+1}$$
Now, $z\bar z=|z|^2=|\cos\theta+i\sin\theta|^2=1$
A: $$
\begin{equation}
\begin{split}
\frac{z-1}{z+1} &= \frac{\cos\theta + i\sin\theta-1}{\cos\theta + i\sin\theta+1} \\
 &= \frac{\cos\theta-1+i\sin\theta}{\cos\theta+1+i\sin\theta} \\
&=\frac{-2\sin^2\frac{\theta}{2} + i\sin\theta}{2\cos^2\frac{\theta}{2}+i\sin\theta} \\
&= \frac{2\sin\frac{\theta}{2}}{2\cos\frac{\theta}{2}}\cdot\frac{-\sin\frac{\theta}{2}+i\cos\frac{\theta}{2}}{\cos\frac{\theta}{2}+i\sin\frac{\theta}{2}} \\
&=\tan\frac{\theta}{2}\cdot i\cdot \frac{i\sin\frac{\theta}{2}+\cos\frac{\theta}{2}}{\cos\frac{\theta}{2}+i\sin\frac{\theta}{2}}\\
&= i\tan\frac{\theta}{2}
\end{split}
\end{equation} \\
$$
Clearly the above expression is purely imaginary.
A: Using $\displaystyle\cos2A=2\cos^2A-1=1-2\sin^2A $ and $\displaystyle\sin2A=2\cos A\sin A,$
$$\frac{\cos\theta+i\sin\theta-1}{\cos\theta+i\sin\theta+1}=\frac{-2\sin^2\frac{\theta}2+i2\sin\frac{\theta}2\cos\frac{\theta}2}{2\cos^2\frac{\theta}2+i2\sin\frac{\theta}2\cos\frac{\theta}2}$$
$$=\frac{2i\sin\frac{\theta}2\left(\cos\frac{\theta}2+i\sin\frac{\theta}2\right)}{2\cos\frac{\theta}2\left(\cos\frac{\theta}2+i\sin\frac{\theta}2\right)}=i\tan\frac{\theta}2$$  as $\displaystyle\cos\frac{\theta}2+i\sin\frac{\theta}2\ne0$
