1)
Prove that
$\displaystyle ||z_1|-|z_2||\leq|z_1-z_2|$
2)
Show that if $\displaystyle |z|<1$
then
$\displaystyle |z-1|+|z+1|\leq2$
First one looks like the triangle inequality but i cant go further.
Hello, Nrt!
I believe there is a typo in #2.
2) Show that if $\displaystyle |z|\:<\: 1$
then: .$\displaystyle |z-1|+|z+1| \;\;{\color{red}\geq}\;\;2$
Since $\displaystyle |z| < 1,\:z$ is in the unit circle centered at the origin.
. . $\displaystyle \begin{array}{c}|z-1|\text{ is its distance from }A(1,0) \\ \\[-4mm] |z+1|\text{ is its distance from }B(\text{-}1,0) \end{array}\bigg\}\quad\text{ endpoints of a diameter, length 2}$
Code:| o o o o | z o o | ∆ o o *| * o * | * o * | * o - B∆ - - - - + - - - - ∆A - o | o | o | o o | o o | o o o o |
From the triangle inequality: .$\displaystyle \overline{zA} + \overline{zB} \:\geq \:\overline{AB}$
. . Therefore: .$\displaystyle |z-1| + |z+1| \:\geq \:2$
It is the exact same proof for complex numbers as for real real.
You must realize that $\displaystyle |z|$ is just a real number.
The triangle inequality holds for complex numbers: $\displaystyle \left| {z + w} \right| \leqslant \left| z \right| + \left| w \right|\;\& \,\left| {z - w} \right| = \left| {w - z} \right|$.
So you can get $\displaystyle - \left| {z - w} \right| \leqslant \left( {\left| z \right| - \left| w \right|} \right) \leqslant \left| {z - w} \right|$.
At this point you have nothing but real numbers and you use this fact.
If $\displaystyle a\ge 0$ and $\displaystyle -a\le b \le a$ then $\displaystyle |b|\le |a|$ or $\displaystyle \left| {\left| z \right| - \left| w \right|} \right| \leqslant \left| {z - w} \right|$.
See you let $\displaystyle a = \left| {z - w} \right|$ and $\displaystyle b = \left| z \right| - \left| w \right|$ noting that $\displaystyle \left| {\left| {z - w} \right|} \right| = \left| {z - w} \right|$.