z⁴+(√3-3i)(1+z)⁴=0.

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.

p²=(psinθ)²+(pcosθ)²=1/16+1/48=1/12, p=±1/√12=±1/(2√3)=±√3/6.

We can replace the complex trig expression by peꜞᶿ.

(1+1/z)⁴=peꜞᶿ.

1+1/z=p⁰⋅²⁵e⁰⋅²⁵ꜞᶿ.

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:

i(⁴√p)e^(iθ)=(⁴√p)e^i(θ+π/2)=(⁴√p)(cos(θ+π/2)+isin(θ+π/2)).