# Upper and Lower Bounds

• Sep 4th 2010, 10:10 AM
dwsmith
Upper and Lower Bounds
Find an upper and lower bound for the reciprocal of the modulus of $z^4-5z^2+6$ if $|z|=2$

$n\leq\frac{1}{|z^4-5z^2+6|}\leq m$

Out of all the the ways to break modulus how do I know which structure of the triangle inequalities to use?

$|z^4-5z^2+6|=|(z^2-3)(z^2-2)|\geq|z^2-3||z^2-2|=||z|^2-|3||*||z^2|-|2||$ $=|4-3||4-2|=1*2=2$

$|z^4-5z^2+6|\leq|z|^4-5|z|^2+6=2^4-5*2^2+6=16-20+6=2$

Why am I getting the same value for my upper and lower bounds?
• Sep 4th 2010, 12:01 PM
HallsofIvy
It is not necessary to do that. A is an upper bound for 1/f(x) if and only if 1/A is a lower bound (not 0) for f(x) and B is a lower bound for 1/f(x) if and only if 1/B is an upper bound for f(x).
• Sep 4th 2010, 12:04 PM
dwsmith
I don't understand since the bounds are $\frac{1}{42}\leq\frac{1}{modulus}\leq\frac{1}{2}$