179. The range of |z| , |z^2+a|=|bz+c|. Particular cases.

by Virgil Nicula, Nov 28, 2010, 9:37 AM

Proposed problem. Ascertain the range of $|z|$ , where $z\in\mathbb C$ for which $\left|z^2+a\right|=|bz+c|$ and $\{a,b,c\}\subset\mathbb R$ are given , $a>0$ . Study only some particular cases.

Proof of the following particular cases. Denote $|z|=\rho$ and $s=z+\overline z$ . Observe that $z\cdot\overline z=\rho^2$ and $s=z+\overline z=2\cdot \mathrm{Re}\ (z)$ . Therefore :

$\blacktriangleright\ \left|z^2+1\right|=|z-1|$ $\iff$ $\left(z^2+1\right)\cdot\left(\overline z^2+1\right)=(z-1)\cdot \left(\overline z-1\right)$ $\iff$ $\rho^4+s^2-2\rho^2+1=\rho^2-s+1$ $\iff$

$\left(\rho^2-\frac 32\right)^2+\left(s+\frac 12\right)^2=\frac 52$ . In conclusion, $\left|\rho^2-\frac 32\right|\le \sqrt {\frac 52}$ $\iff$ $|z|\le\sqrt {\frac {3+\sqrt{10}}{2}}$ with equality if and only if $\mathrm{Re}\ (z)=-\frac 14$ .

$\blacktriangleright\ \left|z^2+1\right|=|4z+3|$ $\iff$ $\left(z^2+1\right)\cdot\left(\overline z^2+1\right)=(4z+3)\cdot \left(4\overline z+3\right)$ $\iff$ $\rho^4+s^2-2\rho^2+1=16\rho^2-12s+9$ $\iff$

$\left(\rho^2-9\right)^2+(s-6)^2=125$ . In conclusion, $\left|\rho^2-9\right|\le 5\sqrt 5$ $\iff$ $|z|\le\sqrt {9+5\sqrt 5}$ with equality if and only if $\mathrm{Re}\ (z)=3$ .

$\blacktriangleright\ \left|z+\frac 1z\right|=a>0\ \ \implies\ \ \frac {-2+\sqrt {a^2+4}}{2}\le |z|\le\frac {2+\sqrt {a^2+4}}{2}$ with equality if and only if $\mathrm{Re}\ (z)=0$ .

$\blacktriangleright\ \left|z^2+1\right|=2\cdot |z+1|$ $\implies$ $|z|\le \sqrt 7$ with equality if and only if $\mathrm{Re}\ (z)=1$ .

An easy extension. Consider $\{a,b\}\subset\mathbb R$ and $z\in \mathbb C$ for which $\left|z^2+1\right|=2\cdot |az+b|$ . Denote $A>0$

so that $A^2=\left(a^2+1\right)\cdot \left(a^2+b^2\right)$ . Prove that $|\mathrm{Re}\ z-ab|\le A$ and $\left|\left(2a^2+1\right)-|z|^2\right|\le 2A$ .
This post has been edited 20 times. Last edited by Virgil Nicula, Dec 1, 2015, 11:13 AM

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