Q:

Minimal polynomial of $i\frac{\sqrt{3}}{2}+\frac{1}{2}$

I'm looking for the minimal polynomial of $\alpha = i\frac{\sqrt{3}}{2}+\frac{1}{2}$ in $\mathbb{Q}[x]$. A polynomial with a root $\alpha$ is 
$$(2(x-\frac{1}{2}))^4-9.$$
A computer algebra system shows me that the polynomial is irreducible. I looking for a way to compute this by hand in an exam. Is there any way to do so without using any irreduciblity criterions which where not part of my lecture?

A:

You can also do this by thinking geometrically - from trigonometry, you might remember that $$e^{i \pi/3} = \frac{1}{2}+ i \frac{\sqrt{3}}{2} .$$  Therefore, this is a zero of $x^3 +1$.  A quick check shows that this is not irreducible, since $-1$ is a root.  Therefore we can factor it:  $x^3 + 1 = (x -1)(x^2 - x + 1)$.  Thus $ \frac{1}{2}+ i \frac{\sqrt{3}}{2} $ is a root of $x^2 - x + 1$.  The minimal polynomial must have degree $\geq 2$ since $ \frac{1}{2}+ i \frac{\sqrt{3}}{2} \notin \mathbb{Q}. $   Therefore $x^2 -x +1$ must be the minimal polynomial.  Alternately, you can check directly that $x^2 -x +1$ is irreducible over $\mathbb{Q}$ since it is degree 2 and its roots are not in $\mathbb{Q}$.

A:

If  $x=i\frac{\sqrt3}{2}+\frac{1}{2}\implies 2x-1=i\sqrt3\implies x^2-x+1=0 $  is the minimal polynomial with integral coefficients.
Another way to address this is to put $i\frac{\sqrt3}{2}+\frac{1}{2}=R(\cos y+i\sin y)=Re^{iy}$ where $R$ is a positive real number.
Squaring & adding we get , $R^2=1\implies R=1$
So, $\cos y=\frac{1}{2}$ and  $\sin y=\frac{\sqrt3}{2}$  $\implies \tan y=\sqrt3$
As the cosine and sine ratios of $y$ are positive, so $y=\frac{\pi}{3}$
So, $x=i\frac{\sqrt3}{2}+\frac{1}{2}=e^{\frac{i\pi}{3}}$ 
To rationalize $e^{\frac{ip\pi}{q}}$ where $p,q$ are integers with (p,q)=1, we need to take the q-th power as $e^{ip\pi}=(-1)^p$
So $x^3=-1\implies x^3+1=0$, but clearly,$x≠-1$
So, $\frac{x^3+1}{x+1}=0\implies x^2-x+1=0$

