IrishBoy123
  • IrishBoy123
complex numbers
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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IrishBoy123
  • IrishBoy123
|z| = 2 prove \[|z^2 - 7 | \ge 3\]
idku
  • idku
why can't you just plug it in?
idku
  • idku
|2| raised to the second power is 4. Or is this something different than alg?

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freckles
  • freckles
z is a complex number a+bi doesn't have to be 2
idku
  • idku
oh, ok.
freckles
  • freckles
just the magnitude of a+bi has to be 2
freckles
  • freckles
\[|z^2-7|=|7-z^2|\ge|7|-|z^2|=|7|-|z|^2=7-4=3\]
IrishBoy123
  • IrishBoy123
et voila! https://gyazo.com/f606fea88651c5a8242fb8ea5f2e9db9 what about the LHS of that?
freckles
  • freckles
\[|z^2-7| =|z^2+(-7)| \le |z^2|+|(-7)|=|z|^2+7=4+7=11\]
freckles
  • freckles
\[3 \le |z^2-7| \le 11\]
freckles
  • freckles
wait did you mean right hand side of that?
freckles
  • freckles
because the first thing was the left hand side of that one thingy
IrishBoy123
  • IrishBoy123
https://gyazo.com/7c7699ae9dad7798832fed473b22ba9d
freckles
  • 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
Loser66
  • Loser66
\[|z^2|=|z^2 -7 +7|\leq |z^2 -7|+|7|\] Hence \(|z^2 -7|\geq|z^2|-|7| \) Therefore \(||z^2 -7||\geq|z^2|-|7||=|4-7|=3\)
freckles
  • 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
Loser66
  • Loser66
are you sleeping? @IrishBoy123 hahaha
freckles
  • freckles
no that is his cat with his bum on the keyboard
IrishBoy123
  • 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.
IrishBoy123
  • IrishBoy123
you two are just mad. sleeping with a bum on my keyboard or something!
freckles
  • freckles
i have no cat butts allowed sign on my keyboard

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