How to extend this inequality? It can be easily verified that for any real $\theta$ and complex number $a$,
$$\Re\left(\frac{e^{i\theta}}{e^{i\theta}-a}\right)\leq\frac12,\quad\text{ for }|a|\geq1.$$
I was trying to verify this result for any complex number $a$ with $0\leq|a|<1$. By intuition, I think the inequality gets this shape:
$$\Re\left(\frac{e^{i\theta}}{e^{i\theta}-a}\right)\leq \frac{1}{2}+\frac{1}{2|e^{i\theta}-a|^2}(1-|a|^{2}).$$
Can we generalise this to two terms as follows?
$$\Re\left(\frac{e^{i\theta}}{e^{i\theta}-a}+\frac{e^{i\theta}}{e^{i\theta}-b}\right)\leq 1+\frac{1}{4}\left|\frac{1}{e^{i\theta}-a}+\frac{1}{e^{i\theta}-b}\right|^2(1-|a|^{4}),$$
where $|a|\leq|b|< 1?$
 A: Assuming $a$ and $\theta$ are real, your intuition is  an equality, not an inequality.
$$\Re\left(\frac{e^{i\theta}}{e^{i\theta} - a}\right) = \frac{1-a \cos(\theta)}{1 - 2 a \cos(\theta) + a^2} = \frac{1}{2} + \frac{1-a^2}{2(1-2 a \cos(\theta) + a^2)}$$
EDIT: If $a = r e^{i\phi}$ with $r > 0$ and $\phi$ real, then
$$ \Re\left( \frac{e^{i\theta}}{e^{i\theta}-a}\right) = \frac{1}{2} + \frac{1-r^2}{2(1 - 2 r \cos(\phi -\theta) + r^2)}$$
A: $\def\e{\mathrm{e}}\def\i{\mathrm{i}}\def\Re{\operatorname{Re}}$As showed by @RobertIsrael, the first inequality always holds since it is an identity. But for the second inequality, if $θ = 0$, $a = \dfrac{1}{2} (1 + \i)$, $b = \dfrac{1}{2} (1 - \i)$, then\begin{gather*}
\Re\left( \frac{\e^{\i θ}}{\e^{\i θ} - a} + \frac{\e^{\i θ}}{\e^{\i θ} - b} \right) = \Re((1 + \i) + (1 - \i)) = 2,\\
1 + \frac{1}{4} \left| \frac{1}{\e^{\i θ} - a} + \frac{1}{\e^{\i θ} - b} \right|^2 (1 - |a|^4) = 1 + \frac{1}{4} |(1 + \i) + (1 - \i)|^2 \left(1 - \left( \frac{1}{\sqrt{2}} \right)^4 \right) = 2 - \frac{1}{4}.
\end{gather*}
Therefore the second inequality does not necessarily hold for $|a| \leqslant |b| < 1$.
