Complex Numbers: Finding solutions to $ (z^2-3z+1)^4 = 1 $ I have to find all solutions to
$$ (z^2-3z+1)^4 = 1 $$
What I thought could work was 
$$z^2-3z+1= 1^{1/4} $$
Given that the 4 4th-roots of 1 are $1, i, -i, -1$ my idea was to look at each case separately. Starting with $1$ and with $z=a+bi \quad a,b \in R$:
$$z^2-3z+1= 1\\z^2-3z=0\\a^2+2abi-b^2=3a+3bi$$
From where you get
$$2ab=3b\ \to \ a=\frac 32\\a^2-b^2=3a\ \to \ \frac94-b^2=\frac92 \ \to \ b\notin R$$
So no possible solutions in this case.
I think the idea is okay but when I try to do the same with $i$:
$$z^2-3z+1= i\to (a^2-b^2-3a+1)+i(2ab-3b)=i $$
And I get:
$$a^2-b^2-3a+1 = 0\\2ab-3b=1 $$
Which can be solved but seems overly complicated...
Is what I've done correct? Any simpler ideas?

Ron Gordon's comment on the first case is exactly the way to go for the real roots of unity: simply factor the expression.  For the two complex roots, the quadratic formula should suffice.  You may find it worthwhile to convert the complex radicands to polar form by
$$x+iy=re^{i\theta},\,\,\,r=\sqrt{x^2+y^2},\,\,\,\theta=\tan^{-1}\left(\frac{y}{x}\right).$$
This will make taking that square root a whole lot easier.  For an example, let's take the second case: z^2-3z+(1+i)=0.  The quadratic formula gives us
$$z=\frac{-(-3)\pm\sqrt{(-3)^2-4(1)(1-i)}}{2(1)}=\frac{3\pm\sqrt{9-(4-4i)}}{2}=\frac{3\pm\sqrt{5+4i}}{2}.$$
The radicand here is $5+4i$ and has magnitude $r=\sqrt{5^2+4^2}=\sqrt{41}$ and argument $\theta=\tan^{-1}\left(\frac{4}{5}\right)$.  So
$$5+4i=\sqrt{41}e^{i\tan^{-1}\left(\frac{4}{5}\right)}.$$
Then the square root is simply
$$\sqrt{5+4i}=(5+4i)^{\frac{1}{2}}=\sqrt[4]{41}e^{\frac{1}{2}i\tan^{-1}\left(\frac{4}{5}\right)}.$$
Substitute back into the quadratic formula to obtain those two roots.
