Identify and sketch the locus ${\mathop{\rm Re}\nolimits} \bigl( {{{z + i} \over {z - i}}} \bigr) = 1$. 
Identify and sketch the locus of the point z on a complex plane which satisties the equation ${\mathop{\rm Re}\nolimits} \left( {{{z + i} \over {z - i}}} \right) = 1$.

I tried to do it but I stuck. Can anyone tell me why.
$$\eqalign{
  & {\mathop{\rm Re}\nolimits} \left( {{{z + i} \over {z - i}}} \right) = 1  \cr 
  & \left| {z + i} \right| = \left| {z - i} \right|  \cr 
  & \left| {x + yi + i} \right| = \left| {x + yi - i} \right|  \cr 
  & \left| {x + i(y + 1)} \right| = \left| {x + j(y - 1)} \right|  \cr 
  & \sqrt {{x^2} + {{(y + 1)}^2}}  = \sqrt {{x^2} + {{(y - 1)}^2}}   \cr 
  & {x^2} + {y^2} + 2y + 1 = {x^2} + {y^2} - 2y + 1  \cr 
  & 4y = 0?? \cr} $$
 A: You can solve in a different way.
Take $a,b \in \mathbb{R}$, $z=a+bi$.
$$Re\left(\dfrac{a+(b+1)i}{a+(b-1)i}\right)=1$$
$$Re\left(\dfrac{(a+(b+1)i)(a-(b-1)i)}{(a+(b-1)i)(a-(b-1)i)}\right)=1, a\neq0$$
$$Re\left(\dfrac{a^2+b^2-1+2ai}{a^2+(b-1)^2}\right)=1, a\neq0$$
$$\dfrac{a^2+b^2-1}{a^2+(b-1)^2}=1, a\neq0$$
$$b^2-1=(b-1)^2, a\neq0$$
$$b=1, a\neq0$$
So $z=a+i$, for any $a \in \mathbb{R}\setminus\{0\}$
Consider the case when $a=0$.
$$Re(\dfrac{(b+1)i}{(b-1)i})\neq1, \forall b \in \mathbb{R}$$
A: Say, $T(z)=\frac{z+i}{z-i}$. Then its a Möbius transformation with inverse $S(z)=i\frac{z+1}{z-1}$.
You have to solve $Re(T(z))=1$, which basically asks that "what is mapped to the line $x=1$ by the map $T(z)$?".
So you apply $S$ on the line. Now $S(1-i)=-2+i,S(1)=\infty, S(1+i)=2+i$. So the line $x=1$ is mapped by $T$ to the line through $(-2+i,\infty,2+i)$ which is the line $y=1$. But the point $i$ on this line is sent to $\infty$.
So the solution set is $\{z:Imz=1\}\setminus{\{i\}}$.
A: Here is another answer, though I personally prefer ChesterX's answer if you are allowed to use Mobius transformations, as they are really the underlying reason why the solution set is a generalized circle (circle/line).
For any $z \in \mathbb{C}$:
  $Re(\frac{z+i}{z-i}) = 1$
  $⇔ \dfrac{z+i}{z-i} + (\dfrac{z+i}{z-i})^* = 2$ [where "$z^*$" denotes "complex conjugate of $z$"]
  $⇔ \dfrac{z+i}{z-i} + \dfrac{z^*-i}{z^*+i} = 2$
  $⇔ \dfrac{(z+i)(z^*+i)+(z-i)(z^*-i)}{(z-i)(z^*+i)} = 2$
  $⇔ \dfrac{2zz^*-2}{zz^*+1+i(z-z^*)} = 2$
  $⇔ 2zz^*-2 = 2(zz^*+1+i(z-z^*)) \wedge zz^*+1+i(z-z^*) \ne 0$
  $⇔ 2i(z-z^*) = -4 \wedge 2zz^*-2 \ne 0$
  $⇔ \frac{z-z^*}{2i} = 1 \wedge |z|^2 \ne 1$
  $⇔ Im(z) = 1 \wedge |z|^2 \ne 1$
  $⇔ Im(z) = 1 \wedge z \ne i$
