Draw z in a complex plane, for which: $Re\left( \frac{1}{z-1} + \frac{1}{z+1}\right)>0$
No idea. Tried it all, but it becomes too complicated to sketch it.
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Solve $Re(\frac{1}{z-1}+\frac{1}{z+1}) = 0$. it should give you $Re(z)= 0$ or $|z|=1$ – reuns Nov 03 '16 at 06:49
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Hint. We have that $$\frac{1}{z-1} + \frac{1}{z+1}=\frac{2z}{z^2-1} =\frac{2z\overline{(z^2-1)}}{|z^2-1|^2}=\frac{2z(\overline{z}^2-1)}{|z^2-1|^2}=\frac{2(|z|^2\overline{z}-z)}{|z^2-1|^2}.$$ Now $$0<\mbox{Re}\left( \frac{1}{z-1} + \frac{1}{z+1}\right) \Leftrightarrow 0<\mbox{Re}(|z|^2\overline{z}-z)=|z|^2\mbox{Re}(\overline{z})-\mbox{Re}(z)=(|z|^2-1)\mbox{Re}(z).$$Can you take it from here?
Robert Z
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The boundary of the domain is formed by the points such that the real part is zero.
$$\frac{2z}{z^2-1}=\frac1{it}$$ (the inverse was used for convenience).
Solving the quadratic equation for $z$,
$$z=\pm\sqrt{1-t^2}+it,$$
which is the circle $x^2+y^2=1$.