Prove $|\alpha-i|=|\beta-i|$ if $\alpha$ and $\beta$ are the roots of $z+\dfrac{1}{z}=2e^{i\theta}$ 
If $\alpha$ and $\beta$ are the roots of $z+\dfrac{1}{z}=2(\cos\theta+i\sin\theta),$ $0<\theta<\pi$, then prove that $|\alpha-i|=|\beta-i|$

My Attempt
$$
z^2-2e^{i\theta}z+1=0\implies\alpha+\beta=2e^{i\theta}\quad\&\quad\alpha\beta=1\\
\alpha,\beta=\frac{2e^{i\theta}\pm\sqrt{4e^{2i\theta}-4}}{2}=e^{i\theta}\pm\sqrt{e^{2i\theta}-1}=e^{i\theta}\pm\sqrt{\cos2\theta-1+i\sin2\theta}\\
=e^{i\theta}\pm\sqrt{-2\sin^2\theta+2i\sin\theta\cos\theta}=e^{i\theta}\pm\sqrt{2\sin\theta}.\sqrt{-\sin\theta+i\cos\theta}\\
=e^{i\theta}\pm\sqrt{2\sin\theta}.\sqrt{\cos(\tfrac{\pi}{2}+\theta)+i\sin(\tfrac{\pi}{2}+\theta)}\\
=e^{i\theta}\pm\sqrt{2\sin\theta}.\Big[{\cos(\tfrac{\pi}{4}+\tfrac{\theta}{2})+i\sin(\tfrac{\pi}{4}+\tfrac{\theta}{2})}\Big]\\
$$
As it was asked as a multiple choice question with the solution being one of the options, I think my attempt seems to be more complicated. So what is suggested to be the easiest way to find the solution ?. Can I use geometry ?
 A: This problem can be worked out using polar coordinates, but it seems to be a bit simpler to use rectangular coordinates.
That $|\alpha-i|=|\beta-i|$ follows from proving this claim:  If $z$ is any root of $z+\frac{1}{z}=2e^{i\theta}$ with $0 \lt \theta \lt \pi$, then $|z-i| = \sqrt{2}$.
Let $z = x + yi$ and let $2e^{i\theta} = a + bi$, so $4 = a^2 + b^2$ and $b \gt 0$.  The quantity $x^2 + y^2$ will appear often, so abbreviate it as $K$.
With these substitutions, $z+\dfrac{1}{z}=2e^{i\theta}$ becomes $x + yi + \dfrac{x - yi}{K} = a + bi$.  Thus $a = x(1 + 1/K)$ and $b = y(1 - 1/K)$.  Substituting these into $4 = a^2 + b^2$, we find
$$
\begin{align}
4 &= x^2(1 + 2/K + 1/K^2) + y^2(1 - 2/K + 1/K^2)\\
&= (x^2 + y^2) + 2(x^2 - y^2)/K + (x^2 + y^2)/K^2\quad\\
&= K + 2(K - 2y^2)/K + 1/K.\\
\end{align}
$$
After multiplying by $K$ and rearranging, this becomes
$$
4y^2 = K^2 - 2K + 1.
$$
Clearly, $2y = \pm(K - 1)$, but which sign is correct?  Since $b = y(1 - 1/K)$, we have $bK = y(K - 1)$.  The left side is positive, so $y$ and $K - 1$ must have the same sign.
Thus we have $2y = x^2 + y^2 - 1$, or $2 = x^2 + (y-1)^2$.  This defines a circle of radius $\sqrt{2}$ centered at $(0,1)$.  In complex terms, the distance from $z$ to $i$ is $\sqrt{2}$, as claimed.
