I'm working on the following problem for my introductory complex variables course.
By factoring $z^4-4z^2+3$ in two quadratic factors and using inequality derived from the triangle inequality, show that if $z$ lies on the circle $|z|=2$, then $$ \bigg|\frac{1}{z^4-4z^2+3}\bigg| \ge\frac{1}{19} $$
I'm not really sure how to attack this problem, I've tried multiple methods but can't seem to get anywhere with them. My first attempt was factoring $z^4-4z^2+3$ into $z^2(z+2)(z-2)+3$ to try and use it with the triangle inequality, but am not really sure how to implement this into the triangle inequality.
A general hint towards solving problems similar to this will suffice. I'm not looking for an exact answer, but any help will be appreciated.
Thanks!
Definitely a typo.
$z^4 - 4z^2 + 3 = (z^2 -1)(z^2 - 3)$. If $|z| = 2$ then $|z^2| = 4$.
$|z^2 - 1| \le |z^2| + 1 = 5; |z^2 -3| \le |z^2| + 3 = 7$ so
$|z^4 - 4z^2 + 3| = |z^2 -1||z^2 -3|\le 5*7 = 35$
And $|\frac 1 {z^4 - 4z^2 + 3}| \ge \frac 1{35}$
Equality holds if $|z^2 - 1| = |z^2| +1$ and $|z^2 - 3| = |z^2| +3$ which happens if $z^2 = -4$ or if $z = \pm 2i$.
So $|\frac 1 {z^4 - 4z^2 + 3}|_{z = \pm 2i} = \frac 1{35} < \frac 1{19}$
On the other hand if $|z^2| = 4$ then $|z^2 - 1| \ge |z^2| - 1 = 3$ and $|z^2 - 3| > |z^2| - 3 = 1$ (with equality holding if $z = \pm 2$) and so $ \frac 1{35} \le |\frac 1 {z^4 - 4z^2 + 3}| \le \frac 13$.