If $|z-\frac 3z|=2$, Find the greatest value of $|z|$

Question

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

And $$|z^2-3|\le |z|^2+3$$ $$2|z|\le |z|^2+3$$ which isn’t a valid equation, what’s going wrong?

Answer 1

You are replacing an equation with an inequation, which is a weaker condition. Doing that, you integrate alien solutions. (Though the inequation is "valid".)

Answer 2

Hint

Use https://www.cut-the-knot.org/arithmetic/algebra/ComplexNumberInequalities.shtml

$$|z|+\dfrac3{|-z|}\ge \left|z-\dfrac3z\right|\ge|z|-\dfrac3{|z|}$$

Answer 3

Your inequality $2|z|\le|z|^2+3$ is very much valid. For example, $|z|=2$ satisfies it.

The maximum value of $|z|$ that allows this inequality is the maximum root of the corresponding equality $2|z|=|z|^2+3$, which can be solved by the usual methods for quadratic equations to give $|z|=3$.

Thus $|z|=3$ is an upper bound. To prove that it is the sharp upper bound, meaning the true maximum, simply try $z=3$ itself in your original equation $|z-(3/z)|=2$.

Answer 4

Maximizing $a^2+b^2$ with the constraint $(a^2+b^2)^2-10a^2+2b^2+9 = 0$ (please see comments above) results in the two solutions $(3,0)$ and $(-3,0)$, thus the greatest value of $z$ is 3 and occurs at $z\in \{ -3,3\}$.

Here is a picture of the curve plotted in Mathematica along with the circle $|z|=3$.