2
$\begingroup$

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!

$\endgroup$
  • 3
    $\begingroup$ The statement is false. Take $z=-2$. Then $\left|\frac1{z^4-4z^3+3}\right|=\frac1{51}<\frac1{19}$. $\endgroup$ – José Carlos Santos May 27 '17 at 18:10
  • 1
    $\begingroup$ Quadratic factors have the form $(az^2+bz+c)$; your factorisation is into a quadratic factor, two linear and a remainder. If you set $w=z^2$ and consider the denominator now as $w^2-4w+3$ you can probably get the right two factors pretty quickly! $\endgroup$ – postmortes May 27 '17 at 18:18
  • 1
    $\begingroup$ Factoring within sums (example $x^2 +4x + 4 = x(x+4) + 4$) almost never reveals anything useful. And $z^2(z+2)(z-2) + 3$ is not "two quadratic factors". Factor the whole thing. $z^4 -4z + 3= (z^2-1)(z^2-3)$ That's two quadratic factors. $\endgroup$ – fleablood May 27 '17 at 18:25
  • 2
    $\begingroup$ The right hand side should be $1/35$, assuming the left hand side is correct. That value is achieved at $z=\pm 2i$. $\endgroup$ – Harald Hanche-Olsen May 27 '17 at 18:29
  • 1
    $\begingroup$ Yeah... I get easily that $(z^2 - 1) \le |z^2| + 1 = 5$ and $|z^2 -3| \le 7$ so the whole thing is greater than $1/35$ and if $z = 2i$ then equality holds and the whole thing is $1/35 < 1/19$. $\endgroup$ – fleablood May 27 '17 at 18:29
1
$\begingroup$

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$.

$\endgroup$
  • $\begingroup$ After working through the problem as everyone above suggested, I did get this answer. Thank you. $\endgroup$ – Kosta May 28 '17 at 1:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.