Question

complex number

Answer 1

for \( |z| = 2\),

prove:

\[|z^2 - 4 z - 3 | \le 15\]

Answer 2

\[-15 \le z^2-4z-3 \le 15 \\ -15+3 \le z^2-4z \le 15+3 \\ -12 \le z^2-4z \le 18 \\ -12+4 \le z^2-4z+4 \le 18+4 \\ -8 \le (z-2)^2 \le 22 \\ 0 \le (z-2)^2 \le 22 \\ (z-2)^2 \le 22 \\ -\sqrt{22} \le z-2 \le \sqrt{22} \\ \]
I wonder if it has anything to do with this
I'm still thinking

Answer 3

I wonder if I take it easy??
Using Triangle in equality, I have
\(|z^2-4z -3|\leq |z^2| +|-4z|+|-3|= |z|^2 +4|z| +|-3)= 2^2 +4*2 +3 =15\)

Answer 4

I think loser's way works awesome

Answer 5

Thank you. I was doubt myself.

Answer 6

@Loser66

which are the sides of the triangle??

Answer 7

can you draw it pls?

Answer 8

That is the triangle inequality theorem.

Answer 9

\(|x+y|\leq |x| +|y|\)

Answer 10

yes, i can see that, it is absolutely marvellous

however, i cannot see how we apply it to \[z^2−4z−3\]

Answer 11

\[|z^2-4z-3|=|z^2+(-4z-3)| \le |z^2|+|(-4z-3)| =|z|^2+|(-1)(4z+3)| \\ =2^2+|-1| \cdot |4z+3|=4+1 \cdot |4z+3| \\= 4+|4z+3| \le 4+|4z|+|3| =4+|4| \cdot |z|+3 \\ =4+4(2)+3=4+8+3=15\]
is it the three term thing @IrishBoy123 ?

Answer 12

Yes! same. :)

Answer 13

http://mathworld.wolfram.com/TriangleInequality.html
triangle inequality can hold for more than two term

Answer 14

yep @freckles it is the 3 term thing

i can see the basic triangle thing

but i am staring at a quadratic in z and it is not so clear

Answer 15

Well you could apply the triangle inequality twice
as I did above

Answer 16

right, i need to read some more

brilliant, @freckles and @Loser66

much appreciated!

i will revert :p

Answer 17

and by read, i mean read 'this thread'

Answer 18

loser was the brilliant one
he was like oh lets use triangle inequality

Answer 19

oops she

Answer 20

I'm so sorry loser

Answer 21

"I'm so sorry loser"

you could not make this up! good night lovely people! sleep well!

Answer 22

well, depends on what time zone you are in, i suppose, but......zzzzzzzzz.

Answer 23

goodnight

Answer 24

got there!


posting a related question. i think i know how to do it now but i am interested in seeing what people have to say.