<s>[INST] If $z$ is a complex number and $Re(z)>1$ then prove that $|\frac{1}{z}-\frac{1}{2}|<\frac{1}{2}$. If $z$ is a complex number and $Re(z)>1$ then prove that $|\frac{1}{z}-\frac{1}{2}|<\frac{1}{2}$.
I tried replacing $z=a+bi$ but it makes problem too long.
 [/INST] You can use the theory of Moebius transformations/inversive geometry to see the behavior of the map $z\mapsto\frac{1}{z}$.  This will tell you that the line $x=1$ will map to a circle.  However, we can do this more explicitly as follows: 
The line $x=1$ has points $1+yi$ on it.  Now, the transformation $z\mapsto\frac{1}{z}$ takes this line to
$$
1+yi\mapsto\frac{1}{1+y^2}-\frac{y}{1+y^2}i.
$$ 
The square of the distance between these points and $\frac{1}{2}$ is given by
$$
\left(\frac{1}{1+y^2}-\frac{1}{2}\right)^2+\left(\frac{y}{1+y^2}\right)^2=\frac{1}{(1+y^2)^2}-\frac{1}{1+y^2}+\frac{1}{4}+\frac{y^2}{(1+y^2)^2}=\frac{1}{4}.
$$
Therefore, the line $x=1$ becomes the circle centered at $\frac{1}{2}$ of radius $\frac{1}{2}$.  Now, all you need to do is to figure out the inside/outside using a test point.
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