# What Coeffecients Allow Me to Factor this Quadratic as $(z-e^{i\phi})(z-e^{i\psi})$

So I have this quadratic of the form:

$$z^2 - \frac{e^{i\theta} + 1}{\overline{a}}z + \frac{a}{\overline{a}}e^{i\theta},$$

where a is in the open unit disk. I'm trying to find values for a and theta which will allow me to factor it into the form $(z-e^{i\phi})(z-e^{i\psi})$ where $e^{i\phi} \neq e^{i\psi}$. My basic approach has been to set:

$$e^{i\phi} + e^{i\psi} = \frac{e^{i\theta} + 1}{\overline{a}},$$

$$e^{i\phi}e^{i\psi} = \frac{a}{\overline{a}}e^{i\theta},$$

and then try to solve for the various variables, but there are just too many unknowns, I can't find a systematic way to do this, hopefully someone can help me with this, thanks.

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If $a=re^{i\alpha}$ then $\overline{a}^{-1}=\frac1re^{i\alpha}$, perhaps that might help? –  bgins Apr 6 '12 at 22:33

Use the quadratic formula : the roots are $$\frac{ \frac{e^{i\theta} +1}{\overline a} \pm \sqrt{ \left( \frac{e^{i\theta}+1}{\overline a} \right)^2 - \frac{4ae^{i\theta}}{\overline a} } }{2}$$ so that you will have two roots of your desired form if and only if those two (one with the $+$ sign, one with the $-$ sign) have norm $1$. In your case, since $a/\overline a$ has norm $1$ and $e^{i\theta}$ too, you know that the product of the roots has norm $1$. Can you find out more?