If the complex number Z satisfies $ |Z^2 - 9| + |Z^2| = 41 $ then the true statements among the following are?

Question

If the complex number Z satisfies $ |Z^2 - 9| + |Z^2| = 41 $ then the true statements among the following are ?

$A)$ $|Z+3| + |Z-3| = 10$
$B)$ $|Z+3| + |Z-3| = 8$
$C)$ Maximum value of $|Z|$ is $5$
$D)$ Maximum value of $|Z|$ is 6

(More than one option may be correct)

The inital equation in the question indicates the fact that the locus of $Z^2$ is an ellipse with foci at $0$ and $9$. But this doesn't help much.

If we put $Z = x + iy$ and simplify. We get an ellipse in $Z$. I get $99y^2 + 63x^2 = 1600$ (but I may be incorrect). How do I proceed from here to validate any of those options?

There seems to exist a much simpler way to solve this question.
All help will be appreciated.

Answer 1

Let $z=Z^2$ and the equation becomes $$ |z-9|+|z|=41 \tag{1}$$ which is an ellipse whose foci are $(0,0)$ and $(9,0)$. Note that $|z|$ reaches it max when $z$ is real and using this, it is easy to see $\max|z|=25$ and hence $\max|Z|=5$. Therefore (C) is the answer.

You also can use the following method: Let $z=x+yi$ and $x=r\cos\theta,y=r\sin\theta$. Then (1) becomes $$ \sqrt{r^2-18r\cos\theta+81} +r=41$$ which imples $$ r^2-r^2-18r\cos\theta+81=(41-r)^2. $$ From this, one has $$ r=\frac{1600}{2(41-9\cos\theta)} \tag{2}.$$ Clearly if $\cos\theta=1$, $r$ reaches the max $5$ or $\max|z|=25$. So $\max|Z|=5$.

Answer 2

The other answer claims C is the only answer, but this is not true. To investigate A and B take the square of $(|z-3|+|z+3|)$ and see what you get:

$$(|z-3| + |z+3|)^2 = |z-3|^2 + |z+3|^2 + 2|z^2-9|$$ since $|z-3|\cdot |z+3| = |(z-3)(z+3)| = |z^2-9|$. Next $|z-3|^2 + |z+3|^2 = 2|z|^2 + 2\cdot 3^2$ by the parallelogram law so

$$(|z-3| + |z+3|)^2 = 2\left(|z^2| + |z^2-9|\right) + 18 = 2\cdot 41 + 18 = 10^2$$

This shows that A is also an answer.