Roots of $z^3 + 3iz^2 + 3z + i = 0$?

Question

I am asked to solve the equation in the title by first solving $\Big(\frac{z+1}{z-1}\Big)^n = i$, which is fine, but I can’t seem to manipulate this equation to get the equation. I’ve tried $n =$ multiples of $3$ or any combinations of $n$, but to no avail. Please advise, thank you!

Answer 1

$(z+1)^3 = -i(3z^2+1) +3z^2+1 = (3z^2+1)(1-i)$

$(z-1)^3 = z^3-3z^2+3z-1 = (z^3+3z) - (3z^2+1)= -i(3z^2+1) - (3z^2+1) = -(1+i)(3z^2+1)$

Thus: $\left(\dfrac{z+1}{z-1}\right)^3= \dfrac{(z+1)^3}{(z-1)^3} = \dfrac{(3z^2+1)(1-i)}{(3z^2+1)(-(1+i))}= \dfrac{-1+i}{1+i}= a+bi$.

Can you find $a, b$ and take cube roots using Demoivre's formula? and this is what the hint tells you to do first.

Answer 2

Note the $\left(\frac{z+1}{z-1}\right)^3 = i \implies (z+1)^3 = i (z-1)^3 \implies (1-i)z^3 + 3 (1+i)z^2 +3(1-i) z + (1+i)=0$. But $1+i = i(1-i)$. Therefore, $$\left(\frac{z+1}{z-1}\right)^3=i \implies (1-i)(z^3+3iz^2+3z+i)=0.$$