Maximum and Minimum value $|z|$, given $|2z + 3i| = |z^2|$

Question

Let $z = a + ib$ (where $a,b \in R$ and $i=\sqrt{-1}$) such that $|2z + 3i| = |z^2|$, Then find the value of $|z|_{max}$ and $|z|_{min}$

My concept, I used $z = a + ib$, hence $|2a+i(2b+3)| = |a^2-b^2+i2ab|$.

On solving I am getting $4a^2+(2b+3)^2 = (a^2+b^2)^2$ , from here I cannot proceed

Answer 1

Making $z = \rho e^{i\phi}$ we have

$$ \rho^2 = \sqrt{4\rho^2+12\rho\sin\phi+9} $$

or

$$ \rho^4 = 4\rho^2+12\rho\sin\phi+9 $$

now making the lagrangian

$$ L(\rho,\phi,\lambda) = \rho-\lambda(4\rho^2+12\rho\sin\phi+9-\rho^4) $$

the stationary conditions give

$$ \nabla L = 0 = \left\{ \begin{array}{l} \lambda \left(4 \rho ^3-8 \rho -12 \sin (\phi )\right)+1 \\ -12 \lambda \rho \cos (\phi ) \\ \rho ^4-4 \rho ^2-12 \sin (\phi ) \rho -9 \\ \end{array} \right. $$

and solving we get at

$$ \left[ \begin{array}{ccc} \rho & \phi & \lambda\\ 1 & -\frac{\pi }{2} & -\frac{1}{8} \\ 3 & \frac{\pi }{2} & -\frac{1}{72} \\ \end{array} \right] $$

NOTE

$$ \rho ^4-4 \rho ^2-12 \rho -9 =(\rho+1)(\rho-3)(\rho^2+2\rho+3)\\ \rho ^4-4 \rho ^2+12 \rho -9 =(\rho-1)(\rho+3)(\rho^2-2\rho+3) $$