Question

If complex number $z(z\ne 2)$ satisfies the equation $z^2=4z+{\left|z\right|}^2+\frac{1}{{\left|z\right|}^3}$, then the value of ${\left|z\right|}^{-4}$ is

Answer 1

$z={re}^{i\theta }\frac{1}{z}=\frac{1}{r}{e}^{-i\theta }$
Substituting in the given equation
$r^2{e}^{2i\theta }=4r{e}^{i\theta }+r^2+\frac{1}{r^3}$

Comparing real part and imaginary part
$r^2\cos 2\theta =4r\cos \theta +r^2+\frac{1}{r^3}$ ...(1)

and

$r^2\sin 2\theta =4r\sin \theta$ ... (2)

Solving (2), we get

$r^2\sin 2\theta =4r\sin \theta$
$\Rightarrow {2r}^2\sin \theta \cos \theta -4r\sin \theta =0$
$\Rightarrow r\sin \theta =0$ or $r\cos \theta =2$

Taking cases
Case I:

$r\sin \theta =0$,

$\Rightarrow \cos \theta =\pm 1$

Use $\cos \theta =1$ in (1)

There is no solution.

Use $\cos \theta =-1$ in (1)
$r^4=\frac{1}{4}$

Case II:

$r\cos \theta =2$

There is no solution.

Therefore
${|z|}^{-4}=r^{-4}=4$