How do I show that the zeroes of $f(z) = z^4 -2\cos(2\theta) z^2 +1$ are $e^{i\theta}$, $e^{-i\theta}$, $-e^{i\theta}$, $-e^{-i\theta}$? N.B. $z$ is a complex number. 
This is what I have so far:
$$f(z) = z^{4} -2 \cos(2\theta) z^2 +1 =R^4 (\cos(4\theta)+i\sin(4\theta))-2R^2\cos(2\theta)(\cos(2\theta)+i\sin(2\theta)) +1 =0$$
$$ = R^4 (\cos(4\theta)+i\sin(4\theta)) -2R^2 \cos^2(2\theta) -2R^2i\cos(2\theta)\sin(2\theta)+1$$
$$=R^4 (\cos(4\theta)+i\sin(4\theta)) -R^2 (\cos(4\theta)+1)-R^2i\sin(4\theta)+1$$
$$=\cos(4\theta)(R^4 - R^2)+i\sin(4\theta)(R^4-R^2)-(R^2-1)$$
$$=(R^2-1)(R^2\cos4\theta +R^2 i \sin(4\theta)-1)$$
$$=(R^2-1)(R^2e^{i4\theta}-1)$$
and that's as far as I can get - if somebody would be able to point out where I've gone wrong, how I can get from here to the answer or a better way of doing this, anything would be much appreciated - thanks!
 A: You can view the equation $z^4-2\cos(2\theta)z^2+1=0$ as a quadratic in the variable $z^2$. Hence by the quadratic formula 
$$z^2=\dfrac{2\cos(2\theta)\pm\sqrt{4\cos^2(2\theta)-4}}{2}=\cos(2\theta)\pm i\sin(2\theta)=e^{\pm2i\theta}.$$ 
Therefore, upon square rooting both sides (consider roots of unity) you obtain the values given above. 
To see the expression for $z^2$ simplifies down sufficiently recall $\cos^2(x)+\sin^2(x)=1$ for all $x\in\mathbb{R}$. 
A: You could always multiply $(z-e^{i\theta})(z+e^{i\theta})(z-e^{-i\theta})(z+e^{-i\theta})$ out and see what you get.
Or, from where you left off, how about writing $R^2e^{4i\theta}-1=(Re^{2i\theta}+1)(Re^{2i\theta}-1)=(\sqrt Re^{i\theta}+1)(\sqrt Re^{i\theta}-1)(\sqrt Re^{i\theta}+i)(\sqrt Re^{i\theta}-i)$.  So in fact, if what you did is correct, we get $\pm\sqrt Re^{i\theta}, \pm i\sqrt Re^{i\theta}$, instead.
Note that you used the same $\theta$ for $z$ as appears in the equation.  You haven't really got a polynomial.  The whole unit circle, $R=1$, is a solution.
