## IrishBoy123 one year ago complex numbers

1. IrishBoy123

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

2. idku

why can't you just plug it in?

3. idku

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

4. freckles

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

5. idku

oh, ok.

6. freckles

just the magnitude of a+bi has to be 2

7. freckles

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

8. IrishBoy123

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

9. freckles

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

10. freckles

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

11. freckles

wait did you mean right hand side of that?

12. freckles

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

13. IrishBoy123
14. freckles

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

15. Loser66

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

16. freckles

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

17. Loser66

are you sleeping? @IrishBoy123 hahaha

18. freckles

no that is his cat with his bum on the keyboard

19. IrishBoy123

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.

20. IrishBoy123

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

21. freckles

i have no cat butts allowed sign on my keyboard