Complex 'show' question (1 Viewer)

gamja · Jan 14, 2023

Let z be a complex number.

Let $w=z+\frac{1}{z}.$

Show that $\left|w^2-3\right|>\left|w\right|^2-3.$

LHS = $\left|z^2+2+\frac{1}{z^2}-3\right|$ = $\left|2cos\left(2\theta \right)-1\right|$ = $4cos^2x-2-1$ = $\left(2cos\theta \right)^2-3$
RHS = $\left|z+\frac{1}{z}\right|^2-3$ = $\left(2cos\theta \right)^2-3$

which is false... where did I go wrong? Thanks for help

tywebb · Jan 14, 2023

2 things are wrong. It should be $\ge$ not > in the question. Also you are assuming |z|=1 which is incorrect.

Anyway you can use triangle ineqality.

$|w^2-3|+|3|\ge|w^2-3+3|$

Therefore $|w^2-3|+3\ge|w|^2$ and so $|w^2-3|\ge|w|^2-3$

------

A simple example showing why you need $\ge$ is if $z=\frac{i-\sqrt3}{2}$ . Then you will see that $w=-\sqrt3$ and so $|w^2-3|$ and $|w|^2-3$ are both 0.

Complex 'show' question (1 Viewer)

gamja

Active Member

tywebb

dangerman

Users Who Are Viewing This Thread (Users: 0, Guests: 1)