if $|z^2-3|=3|z|$ , then the max value of |z| is:

Question

if $|z^2-3|=3|z|$ , then the max value of |z| is:

I attempted this problem by subbing $z$ in as $re^{i\theta}$

so

$$|r^2e^{i2\theta}-3|= 3r$$

$$r=\frac{3}{\sqrt{(\cos2\theta-3)^2+\sin^22\theta}}$$

$$r=\frac{3}{\sqrt{10-6\cos2\theta}}$$

So we just have to find the minimum of the expression below in the denominator to find the maximum value of $r$ which is $|z|$ which is 2 and gives $|z|=3/2$.

On the other hand the author has used the following inequality

$$|z^2-3|-3|z|=0$$

$$|z^2|-3|z|-3<=0$$

Now, i can't figure out why is my answer not the same as the on which we get by solving the quadratic above $(3+\sqrt21)/2$

Answer 1

Your method is wrong. You are treating $r$ and $\theta$ as independent. $\theta$ depends on $r$ so you cannot minimize over all values of $\theta$.

Answer 2

$z=r(\cos t+i\sin t)$ where $r\ge0, t$ is real

$$3r=\sqrt{(r^2\cos2t-3)^2+(r^2\sin2t)^2}$$

As $r\ge0,$ we safely take square on both sides:

$$r^4-3r^2(2\cos2t+3)+9=0$$

Now as $t$ is real, $-1\le\cos2t\le1$

$\implies-1\ge-(2\cos2t+3)\ge-5$

$\implies r^4-3r^2+9\ge r^4-3r^2(2\cos2t+3)+9\ge r^4-15r^2+9$

$\implies \left(r^2-\dfrac32\right)^2+9-\dfrac94\ge0\ge r^4-15r^2+9$

First part is clearly true for real $r$

As the roots of $r^4-15r^2+9=0$ are $\dfrac{15\pm3\sqrt{21}}2$

$0\ge r^4-15r^2+9\implies\dfrac{15-3\sqrt{21}}2\le r^2\le\dfrac{15+3\sqrt{21}}2$

Finally $$\dfrac{15+3\sqrt{21}}2=\dfrac{3^2+(\sqrt{21})^2+2\cdot3\cdot\sqrt{21}}{2^2}=?$$