# Transforming a specific complex function into the form of modulus and argument

Given a specific complex function: $$$$f(t)=\frac{1}{b+c e^{i t}},$$$$

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

$$$$f(t)=\frac{b+c e^{i \varphi (t)}}{b^2-c^2};$$$$

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 $$\varphi (t)$$?

• 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. Oct 29 '19 at 8:07

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.

• No, you don't understand my question. I required that the form should be 𝑟+𝑅*exp[𝑖𝜑(𝑡)], where 𝑟,𝑅 are constants independent of 𝑡. Everyone in the first response comes with your idea. But that is not what I what. Oct 25 '19 at 15:24
• Some more steps attached. Oct 25 '19 at 18:44