Given a specific complex function: \begin{equation} f(t)=\frac{1}{b+c e^{i t}}, \end{equation}
where $b,c,t\in \mathbb{R}$.
Question: I want to express it in terms of the form $r+R e^{i \varphi (t)}$ (which is a circle in the complex plane), where $r,R$ are constants independent of $t$. Then, how?
My preliminary attempt shows that
\begin{equation} f(t)=\frac{b+c e^{i \varphi (t)}}{b^2-c^2}; \end{equation}
however, the phase function $\varphi (t)$ seems to be complicated and so-far no analytic expression is obtained. So, can anyone obtain an analytic expression for $\varphi (t)$?
Hint.
$$ \frac{1}{b+c e^{it}} = \frac{b+c e^{-it}}{(b+c e^{it})(b+c e^{-it})} $$
so we have
$$ \cases{ x = \frac{b}{b^2+2 b c \cos (t)+c^2}+\frac{c \cos (t)}{b^2+2 b c \cos (t)+c^2}\\ y = -\frac{c \sin (t)}{b^2+2 b c \cos (t)+c^2} } $$
now solving for $\sin(t),\cos(t)$ we have
$$ \cases{ \sin(t) = \frac{y (b-c) (b+c)}{c-2 b c x}\\ \cos(t) = \frac{b-x \left(b^2+c^2\right)}{c (2 b x-1)} } $$
and then
$$ \sin^2(t)+\cos^2(t) = 1 $$
after that from the real plane $(x,y)$ to the complex plane is easy.