<s>[INST] Modulus problem (Complex Number) If complex number $z(z\neq2)$ satisfies the equation : $z^2=4z+\lvert z\rvert^2+\frac{16}{\lvert z\rvert^3}$, then what is the value of $\lvert z\rvert^4$?
My try: I tried to take $z=x+iy$ and solved for the value of $\lvert z\rvert^4$ but everytime ended up getting a value greater than $9$.
Hint- The answer lies between $0$ and $9$ , both included.
 [/INST] This question CAN be solved by assuming $z=x+iy$, where $x$ and $y$ are reals.
The given equation can be written as $$z^2-4z= |z|^2+ \frac{16}{|z|^3}$$
Plugging in $z=x+iy$ and 
$$x^2-y^2 +2ixy -4x-4iy=x^2+y^2 + \frac{16}{(\sqrt{x^2+y^2})^3}$$
taking the complex terms to one side, we obtain$$2iy(x-2)=2y^2+4x+\frac{16}{(\sqrt{x^2+y^2})^3}$$
But if $x$ and $y$ are real numbers, then the LHS cannot be equal to the RHS, as one is a real number and the other is a complex number . This means the LHS must be zero.
$\implies y=0$ or $x=2$.
First lets check for $y=0$. The equation reduces to
$$4x+\frac{16}{|x|^3}=0 \implies x=-\sqrt2$$
Hence 
$z=-\sqrt2$
$|z|^4 = 4$
NOTE: $x$ cannot be equal to $2$, because then, the expression can have no real value for $y$.(you can check this yourself)
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