If the point z' is in the half-plane Re(z')>1 , show that…

If the point $z'$ is in the half-plane $\text{Re}(z')>1$, show that the point $z=\dfrac{1}{z'}$ is in the disk $\left|z-\dfrac12\right|<\dfrac12$.

Step 1 · Rewrite the target inequality

$$ \left|\frac1{z'}-\frac12\right| = \left|\frac{2-z'}{2z'}\right| = \frac{|z'-2|}{2|z'|} $$

so we must show $\dfrac{|z'-2|}{2|z'|} < \dfrac12$, i.e. $|z'-2|<|z'|$.

Step 2 · Translate the hypothesis geometrically

With $z'=x'+iy'$, the condition $|z'-2|<|z'|$ becomes:

$$ (x'-2)^2+(y')^2 < (x')^2+(y')^2 \quad\Longrightarrow\quad -4x'+4<0 \quad\Longrightarrow\quad x'>1 $$

which is exactly the hypothesis $\text{Re}(z')>1$: "$z'$ is strictly closer to $2$ than to the origin" is equivalent to lying to the right of the perpendicular bisector of $[0,2]$, i.e. $\text{Re}(z')>1$.

Step 3 · Conclude

Since $\text{Re}(z')>1 \iff |z'-2|<|z'|$, and $z'\neq0$ (as $\text{Re}(z')>1>0$):

$$ \left|\frac1{z'}-\frac12\right| = \frac{|z'-2|}{2|z'|} < \frac{|z'|}{2|z'|} = \frac12 $$

$$ \boxed{\left|z-\frac12\right| < \frac12} $$

Step 4 · Direct Cartesian check (alternative route)

With $z'=x'+iy'$, $x'>1$: $\ z=\dfrac1{z'}=\dfrac{x'-iy'}{(x')^2+(y')^2}$, so $x=\dfrac{x'}{r^2}$, $y=\dfrac{-y'}{r^2}$ where $r^2=(x')^2+(y')^2$. Then:

$$ \left(x-\frac12\right)^2+y^2 = \frac{1-x'}{r^2}+\frac14 $$

Since $x'>1$ and $r^2>0$, $\dfrac{1-x'}{r^2}<0$, so this sum is $<\dfrac14$ — the same conclusion.

Step 5 · Check some key points