Find the maximum value of $| z |$ in $\left| \frac{6z-i}{2+3iz} \right| \leq 1$?

Question

At first I thought that the equation after expanding it out will turn out to be an ellipse or something and I hoped a geometric approach might work, but now I don't think it's that easy. If I haven't made a mistake, we obtain:

$$\left|6(x,y) - (0,1)\right| ^2 \leq \left| 2 + 3i (x,y) \right|^2$$ $$(6x)^2 + (6y-1)^2 \leq (2 - 3y)^2 + (3x)^2$$

I can't see a nice equation coming out of this. I think Lagrange multipliers might work, but doing the calculations will take a while. This is from a standardized exam and it has to be done quickly (i.e. under 3 minutes preferably).

Is there a simple argument using complex numbers that I'm missing?

Answer 1

Assuming that your computations are correct, you can further simplify:

\begin{align} & (6x)^2 + (6y-1)^2 \leq (2 - 3y)^2 + (3x)^2 \\ \implies & 27x^2+27y^2 \leq 3 \\ \implies & x^2+y^2 \leq \frac 19 \end{align}

Hence the answer is $\frac 13$.

And I think this is completely reasonable to do within $3$ minutes.

Answer 2

Note

$$| 6z-i| \leq |2+3iz|$$

or $$(6z-i)(6\bar z+i) \leq (2+3iz)(2-3i\bar z)$$ $$36|z|^2 -i6(\bar z- z) + 1\leq 4 -i6(\bar z- z) + 9|z|^2 $$ $$|z|^2\leq \frac 19 $$

Thus, $|z| \le \frac13$.