If $|z^2-3|=3|z| $ then what is the maximum value of $|z| $ where $z$ is a complex number?

Question

I used triangle inequality like this $$|z^2-3|≤|z|^2+|-3|$$ $$3|z|≤|z|^2+3$$ $$0≤|z|^2-3|z|+3$$ And now let's take $|z|= w$ where $w≥0$ $$0≤w^2-3w+3$$ $$Discriminant=(-3)^2-12=-3$$ $$D<0$$ Hence the inequality is true for all real w But $w≥0$ so it's true for all real non negative $|z|$ So how can there be a maximum value? On the other hand if i use $$||z|^2-|-3||≤|z^2-3|$$ $$||z|^2-3|≤3|z|$$ I get $$|z|^2-|-3|≤3|z|$$ or $$|z|^2-|-3|≥-3|z|$$ The first one gives max value as $(3+✓21)/2$ The second one gives $(-3+✓21)/2$

The correct answer is $(3+✓21)/2$ But I want to know why the other two inequalities got the wrong domains. I don't want to know how to get the answer, i want to know why the two of the inequalities are wrong.

Answer 1

There's nothing wrong about any of the inequalities.

What's happened with the triangle inequality is that it doesn't help any in this case since it tells you what you knew before, namely that $|z|\ge 0.$

As for the third inequality, you made a mistake somewhere. Recall that it gives us the solution $$|z|\le \frac{-3-\sqrt{21}}{2}$$ or $$|z|\ge \frac{-3+\sqrt{21}}{2}.$$ So in this case we cannot say anything about a maximum value. So this doesn't help either.

That leaves the second inequality, which gave you a maximum value.

Answer 2

Solve $\max_{r, \theta} \{ r | |r^2 e^{i2 \theta} -3| = 3r, |\theta| \le \pi \}$ (it is straightforward to check that the constraint set is compact, so a solution exists).

Show that $|r^2 e^{i2 \theta} -3| = 3r$ is equivalent to $(r^2-3)^2 = r^2 ( 3+6 \cos (2 \theta))$ and $r \ge 0$.

It is clear that $r>0$ as it does not satisfy the equality so we can write the constraints as $(r-{3 \over r})^2 = (3 +6 \cos 2 \theta)$, $r>0$. Noting that $r \mapsto (r-{3 \over r})^2$ is increasing for $r \ge \sqrt{3}$, we see that the $\max$ $r>0$ satisfies $(r-{3 \over r})^2 = 9$, or $r^2-3r-3 = 0$. Solving gives $r={1 \over 2} (3+\sqrt{21})$.

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Let $F = \{ z |\ |z^2-3| = 3|z| \}$.

Let $z \in F$ then $z$ satisfies $3|z|\le |z|^2+3$. However, $\{ z |\ 3|z|\le |z|^2+3 \} = \mathbb{C}$.

So the first inequality you derive just states that $F \subset \mathbb{C}$.

If $z \in F$ then $z$ satisfies $|z|^2-3|z|-3 \le 0$ and so we see that $|z| \le {1\over 2} (3 + \sqrt{21})$ and since the right hand side satisfies the constraint, we see that $z^*={1\over 2} (3 + \sqrt{21})$ is a solution.

If $z \in F$ then $z$ satisfies $|z|^2+3|z|-3 \ge 0$ and this just tells us that $|z| \ge {1\over 2} (-3 + \sqrt{21})$.

There is no contradiction.