If a complex number $Z$ satisfies the equation $$|Z^2-9| +|Z^2|=41$$ then prove that locus of $Z$ is an ellipse with centre at origin in the complex plane.
My approach:
One way of solving the question is setting $\,z=x+iy,\, $ but squaring $z$ and squaring it two more times is very lengthy.
So I am guessing that it should have a neat geometric intuition which I can't figure out. Kindly suggest an alternative approach to this question.
Denote $$g(z)=|z^2-9| +|z^2|=41,\quad z \in \mathbb{C}.$$ Clearly, $g(-z)=g(z)=g(\bar z).$ Therefore, the locus $\mathcal{L}(z)$ we are looking for is symmetric over origin and also over real and imaginary axes.
From $41=16+25=25+16$ we can guess solutions $5,-5,4i,-4i.$
Their position in the plane and the symmetry of $\mathcal{L}(z)$ make me think of an ellipse with center at origin, axes $2a=10$ along real axis and $2b=8$ along imaginary axis.
Foci of this ellipse are in $-3$ and $3,$ $(c^2=a^2-b^2,$ where $c$ is the distance between the center and a focus) and its equation in $\mathbb{C}$ is $$|z+3|+|z-3|=10.\tag{1}$$ Let us prove that $(1)$ is an equation of $\mathcal{L}(z).$
Squaring $(1)$ gives an equivalent equation, because all terms in $(1)$ are non-negative. Due to Apollonius theorem we get $$\underbrace{|z+3|^2+|z-3|^2}_{2(|z|^2+9)}+2\cdot\underbrace{|z+3|\cdot |z-3|}_{|(z+3)(z-3)|}=100,$$ or $$|z^2-9|+|z^2|=41.$$