# find maximum and minimum value of $|z|$

If $a,b,c$ are complex number of equal magnitude and satisfy $az^2+bz+c=0,$

then finding maximum and minimum value of $|z|$

with the help of triangle inequality $|az^2+bz+c|\leq |az^2|+|bz|+|c|=|a||z|^2+|b||z|+|c|$

now let $|a|=|b| = |c| = k>0$

so $|az^2+bz+c|\leq k(|z|^2+|z|+1)$

so $|z|^2+|z|+1\geq 0$

wan,t be able to go after that, help me

## 2 Answers

Hints:

• for the maximum value: $\;a z^2 = -bz - c \implies |z|^2 \le |z|+1 \implies |z| \le \cfrac{1+\sqrt{5}}{2}\,$;

• for the minimum value, rewrite the equation as $\;c\cfrac{1}{z^2}+b\;\cfrac{1}{z}+a=0\;$ and use the previous result to show that $\;\cfrac{1}{|z|} \le \cfrac{1+\sqrt{5}}{2} \;\iff\; |z| \ge \cfrac{\sqrt{5}-1}{2}\,$.

We can divide across by $a$ and get $z^2+bz+c= 0$ with $|b|=|c| = 1$. Solving the quadratic gives $z = {1 \over 2} (-b \pm \sqrt{b^2 -4c} )$. From this we get $\sqrt{5}-1 \le 2 |z| \le \sqrt{5}+1$, and by choosing $b=1,c=-1$ we see that these bounds are attained.