Complex 'show' question II (1 Viewer)

gamja · Jan 17, 2023

tywebb said:
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 $|w^2-3|\ge|w|^2-3$

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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.

It is known that $z^{3}+\frac{1}{z^{3}}=w^{3}-3w$ .

from this and the quoted solution above ( $|w^{2}-3|\ge|w|^{2}-3$ ), prove by contradiction that $|z^{3}+\frac{1}{z^{3}}|\le2\ \to|z+\frac{1}{z}|\le2$ .

Assuming that $|w^{3}-3w|\le2\ \to|w|\le2$ to be true,
LHS: $|w||w^{2}-3|\le2$ = $|w|\left(|w|^{2}-3\right)\le2$

and then i need to incorporate the RHS to make a contradiction... Please help!

Thank you

tywebb · Jan 18, 2023

I'd be more inclined to do it by contrapositive

$\begin{aligned}\left|z+\frac{1}{z}\right|>2&\Rightarrow|w|>2\\ &\Rightarrow |w^2-3|\ge|w|^2-3>4-3\\ &\Rightarrow |w||w^2-3|>2\times1\\ &\Rightarrow|w^3-3w|>2\\ &\Rightarrow\left| z^3+\frac{1}{z^3}\right|>2\end{aligned}$

$\text{Hence }\left| z^3+\frac{1}{z^3}\right|\le2\Rightarrow\left| z+\frac{1}{z}\right|\le2$

This is a neater way to do it than by contradiction.

However it can be reconstructed as a proof by contradiction by saying

$\text{Assume }\left| z^3+\frac{1}{z^3}\right|\le2\text{ and }\left| z+\frac{1}{z}\right|>2.$

$\text{Then in the way shown above, }\left| z^3+\frac{1}{z^3}\right|>2\text{ - contradiction.}$

gamja · Jan 18, 2023

Oh I get is now - i did the contradiction proof wrong by assuming the wrong statement

thanks tywebb

Complex 'show' question II (1 Viewer)

gamja

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tywebb

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