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.
 A: Hint:
Let $z=a+ib$ then
\begin{align}
\left|\dfrac{1}{z}-\dfrac{1}{2}\right|<\dfrac12
&\iff \left|\dfrac{1}{a+ib}-\dfrac{1}{2}\right|<\dfrac12\\
&\iff |(2-a)-ib|<|a+ib|\\
&\iff -4a+4<0\\
&\iff a>1
\end{align}
A: Hint:


*

*$\frac{1}{z}-\frac12=\frac{2-z}{2z}$

*$\left|\frac{v}{w}\right| = \frac{|v|}{|w|}$


Using these two facts, you can transform your original inequality into something easier to prove.
A: Hint:
$$
\begin{align}
\left|\,\frac1z-\frac12\,\right|
&=\left|\,\frac{2-z}{2z}\,\right|
\end{align}
$$
$\operatorname{Re}(z)=1$ is the line where points are equidistant from $0$ and $2$.
If $\operatorname{Re}(z)\gt1$, then $z$ is closer to $2$ than to $0$; that is $|z-2|\lt|z-0|$.
A: 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.
