# Complex Numbers and the Triangle Inequality

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!

• The statement is false. Take $z=-2$. Then $\left|\frac1{z^4-4z^3+3}\right|=\frac1{51}<\frac1{19}$. – José Carlos Santos May 27 '17 at 18:10
• 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! – postmortes May 27 '17 at 18:18
• 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. – fleablood May 27 '17 at 18:25
• The right hand side should be $1/35$, assuming the left hand side is correct. That value is achieved at $z=\pm 2i$. – Harald Hanche-Olsen May 27 '17 at 18:29
• 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 $1/35 < 1/19$. – fleablood May 27 '17 at 18:29

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 $\frac 1{35} \le |\frac 1 {z^4 - 4z^2 + 3}| \le \frac 13$.

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