Let $\alpha$ and $\beta$ be the roots of $$x+\frac{1}{x}=2e^{i\phi},0<\phi<\pi.$$
(a) Show that $\alpha+i$ and $\beta+i$ have the same argument and $|\alpha-i|=|\beta-i|$.
(b) Find the locus of the roots $\alpha$ and $\beta$ as $\phi$ varies from $0$ to $2\pi$.
Progress: By symmetry if $t$ is a root then $\frac{1}{t}$ is also a root. So I know that $\alpha=\frac{1}{\beta}$. Moreover we have $$|x+\frac{1}{x}|=2.$$
Formally we obtain $$x=e^{i\phi}\pm\sqrt{e^{2i\phi}-1}\qquad(0<\phi<\pi)\ .$$ Assume $0<\phi<{\pi\over2}$ for the moment, and draw a little figure. It shows that $|e^{2i\phi}-1|=2\sin\phi$ and ${\rm Arg}(e^{2i\phi}-1)={\pi\over2}+\phi$. It follows that $$\pm\sqrt{e^{2i\phi}-1}=\pm\sqrt{2\sin\phi}\>e^{i(\pi/4+\phi/2)}=:\pm w\ .$$ It is easily checked that $w^2=e^{2i\phi}-1$ for arbitrary $\phi\in[0,\pi]$. The two solutions $\alpha$ and $\beta$ therefore are $$\eqalign{\alpha&=\cos\phi+\sqrt{2\sin\phi}\>\cos\left({\pi\over4}+{\phi\over2}\right)+i\left(\sin\phi+\sqrt{2\sin\phi}\>\sin\left({\pi\over4}+{\phi\over2}\right)\right)\ ,\cr \beta&=\cos\phi-\sqrt{2\sin\phi}\>\cos\left({\pi\over4}+{\phi\over2}\right)+i\left(\sin\phi-\sqrt{2\sin\phi}\>\sin\left({\pi\over4}+{\phi\over2}\right)\right)\ .\cr}\tag{1}$$ Plotting these points indicates that they are lying on the circle with center $i$ and radius $\sqrt{2}$. Performing the computation using $(1)$ confirms that indeed $|\alpha-i|^2=|\beta-i|^2=2$ for all $\phi$. It should be possible to obtain this result from the given equation in a simpler way.