Determine the real part of $\left(\cfrac{z-1}{z+1}\right)$ Determine the $Re\left(\cfrac{z-1}{z+1}\right)$ if $z = cos\theta + i \, sin\theta$.
I'm not quite sure whether the right approach would be to stick with the polar form and substitute it into $\left(\cfrac{z-1}{z+1}\right)$ and perhaps use $ 1 = sin^2(\theta) + cos^2(\theta) $. Or maybe I should use the Euler's form.
 A: Multiplying both the numerator and denumerator by $\bar{z}+1$, we get:
$$\frac{z-1}{z+1}\frac{\bar{z}+1}{\bar{z}+1} = \frac{\bar{z}z-1+z-\bar{z}}{z\bar{z}+z+\bar{z}+1}$$
Using $z\bar{z} = |z|^{2}$, $z+\bar{z} = 2\mbox{Re}(z)$ and $z-\bar{z} = 2i\mbox{Im}(z)$ we get:
$$\frac{z-1}{z+1} = \frac{|z|^{2}-1+2i\mbox{Im}(z)}{|z|^{2}+2\mbox{Re}(z)+1}$$
Now you plug $z = \cos\theta + i\sin\theta$ to conclude the exercise.
A: $f(z)=(z-1)/(z+1)$ is a linear fractional transformation. Linear fractional transformations map circles to circles, where straight lines qualify as circles of infinite radius. Since $f(-1)=\infty$, the image of the unit circle is a straight line. Since $f(1)=0$ and $f(i)=(i-1)/(i+1)=i(1+i)/(i+1)=i$, the straight line is the imaginary axis. Hence $\Re(f(z))=0$ for $z=\cos\theta +i\sin\theta$ .
If you don't know about linear fractional transformations, then note that
$$f(e^{i\theta})={e^{i\theta}-1\over e^{i\theta}+1}={e^{i\theta/2}-e^{-i\theta/2}\over e^{i\theta/2}+e^{-i\theta/2}}={2i\sin(\theta/2)\over2\cos(\theta/2)}=i\tan(\theta/2)$$
A: $\frac{z-1}{z+1}=\frac{(z-1)(\bar{z}+1)}{|z+1|^2}$
A: hint
$$\frac{z-1}{z+1}=\frac{z+1-2}{z+1}$$
$$=1-\frac{2}{z+1}$$
$$z+1=(1+\cos(\theta))+i\sin(\theta)$$
$$=2\cos^2(\frac{\theta}{2})+2i\sin(\frac{\theta}{2})\cos(\frac{\theta}{2})$$
$$=2\cos(\frac{\theta}{2})\Bigl(\cos(\frac{\theta}{2})+i\sin(\frac{\theta}{2})\Bigr)$$
$$=2\cos(\frac{\theta}{2})e^{i\frac{\theta}{2}}$$
$$\frac{1}{z+1}=\frac{1}{2\cos(\frac{\theta}{2})}e^{-i\frac{\theta}{2}}$$
$$=\frac 12-\frac{i}{2}\tan(\frac{\theta}{2})$$
The real part you look for is $1-1=0$.
A: Note that $z=\cos\theta + i\sin\theta$ implies $|z|=1$. Now,
$$\frac{z-1}{z+1} = \frac{z\bar z-\bar z}{z\bar z + \bar z} = \frac{1-\bar z}{1+\bar z} = - \overline{ \left(\frac{z-1}{z+1}\right)},$$
so, $\frac{z-1}{z+1}$ is purely imaginary, i.e, the real part is $0$.
A: On the unit circle $z=1/\overline z$ where the overline in complex conjugation.  So
$\dfrac{z-1}{z+1}=\dfrac{1/\overline z-1}{1/\overline z+1}=\dfrac{1-\overline z}{1+\overline z}=-\dfrac{\overline z-1}{\overline z+1}$
Thus $\dfrac{z-1}{z+1}$ is the negative of its own complex conjugate, therefore pure imaginary.  The real part is zero.
