Prove that:
$|z-1|\leq|\sqrt{z^2-1}|<|z+1|$ for R(z)>0
I tried using the general triangle inequality $|z_1|-|z_2|\leq |z_1+z_2|\leq|z_1|+|z_2|$, but I can't seem to get this working.
I would appreciate any help.
Prove that:
$|z-1|\leq|\sqrt{z^2-1}|<|z+1|$ for R(z)>0
I tried using the general triangle inequality $|z_1|-|z_2|\leq |z_1+z_2|\leq|z_1|+|z_2|$, but I can't seem to get this working.
I would appreciate any help.
First note that if $\operatorname{Re}(z) > 0$ then $|z-1| \lt |z+1|$.
This is obvious geometrically, since all points in the right half-plane $x > 0$ are closer to point $(1,0)$ on the $x$ axis than they are to its symmetric across the $y$ axis $(-1,0$). For a formal proof:
$$ \begin{align} \operatorname{Re}(z) > 0 & \iff z + \overline z \gt 0 \\ & \iff z \overline z - (z + \overline z) + 1 \lt z \overline z + (z + \overline z) + 1 \\ & \iff (z-1)(\overline z - 1) \lt (z+1)(\overline z + 1) \\ & \iff |z-1|^2 \lt |z+1|^2 \end{align} $$
When $z \ne 1$ the above gives the following inequalities, which imply the ones to prove:
$$ \begin{align} \left| \frac{z-1}{z+1}\right| \lt 1 \lt \left| \frac{z+1}{z-1} \right| & \quad\iff\quad |z-1|^2 \lt |z^2-1| \lt |z+1|^2 \end{align} $$
Finally, when $z=1$ the given inequalities trivially hold, with equality for the LHS one.