Divide through by z⁴:
1+(√3-3i)(1+1/z)⁴=0, so (1+1/z)⁴=-1/(√3-3i)=-(√3+3i)/12 after rationalisation.
The right-hand side can be written pcosθ+ipsinθ where pcosθ=-√3/12 and psinθ=-¼.
So tanθ=psinθ/(pcosθ)=(-¼)/(-√3/12)=√3, and θ=π/3.
We can replace the complex trig expression by peꜞᶿ.
From this we get z=1/(p⁰⋅²⁵eꜞᶿ-1). Use p=√3/6 and θ=π/12.
This is one solution. We have to consider all the fourth roots of p. Square root of p is ±√p, so the square root of this is ±⁴√p, ±i(⁴√p).
For example, if p=16, square roots would be -4 and +4; the square roots of these are -2i, +2i, -2 and +2.
Another point to remember is that if x=π/2, eꜞˣ=i. This is the same as a phase shift of π/2 in θ. For example: