Complex Expression is Real 
If I have the a complex number $z \in \mathbb{C}$ with absolute value $|z| = 1$, how do I show that $-i \frac {z-1}{z+1}$ is real? 

 A: Note that $$-i\frac{e^{i\theta}-1}{e^{i\theta}+1}\times \frac{e^{-i\theta/2}}{e^{-i\theta/2}}=-i\frac{e^{i\theta/2}-e^{-i\theta/2}}{e^{i\theta/2}+e^{-i\theta/2}}=\tan(\theta/2).$$
A: If $\lvert z \rvert = 1$ then $\overline{z}=z^{-1}$. So complex cojugation of your expression results in $$i\frac{z^{-1}-1}{z^{-1}+1} = -i \frac{z-1}{z+1}.$$
A: Well $\overline{z + 1} =Re(z+1) - iIm(z+1) = (Re z + 1) - iIm(z) = Re(z) - iIm(z) + 1 = \overline z + 1$ for all complex numbers.
SO
$-i \frac {z-1}{z + 1}= -i\frac {(z-1)(\overline z +1)}{(z+1)\overline{(z+1)}}$
$=-i \frac {z\overline z -\overline z + z - 1}{|z + 1|^2}$
$= -i \frac {|z|^2 +i2Im(z) - 1}{|z + 1|^2}$
$=-i \frac {i2Im(z)}{|z+1|^2} = \frac {2Im(z)}{|z+1|^2}$ which is real.
A: A number $w$ is real iff $\operatorname{Im}(w) = \frac{1}{2i}(w - \bar w) = 0\,$. In this case:
$$\require{cancel}
\begin{align}
-i \frac {z-1}{z+1} - \overline{\left(-i \frac {z-1}{z+1}\right)} &= -i \frac {z-1}{z+1} - \left(+i \frac{\bar z - 1}{\bar z + 1}\right) \\
 &= -i \cdot\frac{(z-1)(\bar z + 1)+ (\bar z -1)(z+1)}{(z+1)(\bar z + 1)} \\
 &= -i \frac{|z|^2+\bcancel{z}-\cancel{\bar z}-1+|z|^2+\cancel{\bar z} - \bcancel{z} - 1}{|z+1|^2} = \\
 &= -2i \frac{\cancel{|z|^2}-\cancel{1}}{|z+1|^2} \\
 &= 0
\end{align}
$$
A: A geometric reason is that $1$ and $-1$ are opposite ends of a diameter of the unit circle, and $z$ is on the circumference.
This implies that $z,$ $1,$ and $-1$ are vertices of a right triangle, with the right angle at $z.$
But $z-1$ and $z+1$ have the same lengths and directions as the legs of that triangle;
I’m particular, the difference in their directions is $\frac\pi2$ (a right angle).
Hence the ratio of these two numbers has argument  $\frac\pi2,$ that is, it is purely imaginary, and multiplication by $-i$ produces a real number. 

For an algebraic approach, the complex conjugate of $z+1$ is $z^*+1$ ($1$ added to the conjugate of $z$).
Multiply numerator and denominator by this same amount to get a real number in the denominator.
Use the facts that $zz^*=\lvert z\rvert^2$ and that $z-z^*$ is purely imaginary.
