complex numbers

Question

complex numbers

irishboy123 · Answer

|z| = 2

prove

$$|z^2 - 7 | \ge 3$$

idku · Answer

why can't you just plug it in?

idku · Answer

|2| raised to the second power is 4. Or is this something different than alg?

freckles · Answer

z is a complex number
a+bi doesn't have to be 2

idku · Answer

oh, ok.

freckles · Answer

just the magnitude of a+bi has to be 2

freckles · Answer

$$|z^2-7|=|7-z^2|\ge|7|-|z^2|=|7|-|z|^2=7-4=3$$

irishboy123 · Answer

et voila! https://gyazo.com/f606fea88651c5a8242fb8ea5f2e9db9 what about the LHS of that?

freckles · Answer

$$|z^2-7| =|z^2+(-7)| \le |z^2|+|(-7)|=|z|^2+7=4+7=11$$

freckles · Answer

$$3 \le |z^2-7| \le 11$$

freckles · Answer

wait did you mean right hand side of that?

freckles · Answer

because the first thing was the left hand side of that one thingy

irishboy123 · Answer

https://gyazo.com/7c7699ae9dad7798832fed473b22ba9d

freckles · Answer

$$3=7-4=7-2^2=7-|z|^2=7-|z^2|=|7|-|z^2| \le |7-z^2|=|z^2-7| \ \le |z^2|+|-7|=|z|^2+7=2^2+7=4+7=11$$
this is both left and right hand sides

loser66 · Answer

$$|z^2|=|z^2 -7 +7|\leq |z^2 -7|+|7|$$
Hence $|z^2 -7|\geq|z^2|-|7| $
Therefore $||z^2 -7||\geq|z^2|-|7||=|4-7|=3$

freckles · Answer

$$|7|-|z^2| \le |7-z^2| \le |7|+|-z^2| \ 7-|z|^2 \le |7-z^2| \le 7+|z|^2 \ 7-2^2 \le |7-z^2| \le 7+2^2 \ 3 \le |7-z^2| \le 11 \ 3 \le |z^2-7| \le 11 $$
just in case that one thing was hard to read

loser66 · Answer

are you sleeping? @IrishBoy123  hahaha

freckles · Answer

no that is his cat with his bum on the keyboard

irishboy123 · Answer

that was even more epic!

i'm gonna another one in a new thread.  i can do them now

but then i can post a final question that will connect them all up.

irishboy123 · Answer

you two are just mad.

sleeping with a bum on my keyboard or something!

freckles · Answer

i have no cat butts allowed sign on my keyboard