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?
 A: Starting from $(z^2 - 3z + 1)^4 = 1$, we get 
$$((z^2 - 3z + 1)^2 - 1)((z^2 - 3z + 1)^2 + 1) = 0.$$
For the first term we have that
$$(z^2 - 3z + 1)^2 - 1 = 0.$$
Hence, we get first $4$ roots, $z = 0, 1, 2, 3.$
Next, $$(z^2 - 3z + 1)^2 + 1 = 0$$
$$z^2 - 3z + 1 = i, -i.$$
Using the formula, we get the remaining roots, that is,
$$z = \frac{1}{2}( 3 \pm \sqrt{5 \pm 4i} ).$$
A: 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.
