1
$\begingroup$

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

$\endgroup$
2
$\begingroup$

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 $\;\cfrac{1}{|z|} \le \cfrac{1+\sqrt{5}}{2} \;\iff\; |z| \ge \cfrac{\sqrt{5}-1}{2}\,$.

$\endgroup$
3
$\begingroup$

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.

$\endgroup$

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.