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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 $\varphi (t)$?

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  • $\begingroup$ I think what Cesaro described in his answer is the best you can do. The form you want describes a circle as you said yourself but your function $f$ does not describe a circle in the complex plane so you can't represent it as one. $\endgroup$
    – quarague
    Oct 29 '19 at 8:07
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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.

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  • $\begingroup$ No, you don't understand my question. I required that the form should be ๐‘Ÿ+๐‘…*exp[๐‘–๐œ‘(๐‘ก)], where ๐‘Ÿ,๐‘… are constants independent of ๐‘ก. Everyone in the first response comes with your idea. But that is not what I what. $\endgroup$
    – Mathieu
    Oct 25 '19 at 15:24
  • $\begingroup$ Some more steps attached. $\endgroup$
    – Cesareo
    Oct 25 '19 at 18:44

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