I've this function
$$f(k,\theta) = \frac{1}{k}\frac{1}{\cot\delta_0(k) -i }$$
and i know that $k\cot\delta_0(k) = -\frac{1}{a} + \frac{1}{2}r_ek^2 + \cdots$ it is an expansion.
How can i get that the $\Re{f(k,\theta)} = -a + a^2(a-\frac{1}{2}r_e)k^2$ ?
You probably already solved this by yourself, but to close off this question with an answer, I followed through on my comment.
$$ \Re f(k,\theta) = \frac{1}{k}\frac{\cot \delta_0(k)}{\cot^2 \delta_0(k) +1} $$ multiplying top and bottom by $k^2$ we obtain $$ \Re f(k,\theta) =\frac{k\cot \delta_0(k)}{k^2\cot^2 \delta_0(k) +k^2} $$ using the expansion for $k\cot \delta_0(k) = -\frac{1}{a} + \frac{1}{2}r_ek^2$ the denominator becomes $$ k^2\cot^2 \delta_0(k) +k^2 = \frac{1}{a^2} - \frac{1}{a}r_ek^2 + k^2 = \frac{1}{a^2} +\left(1 -\frac{r_e}{a}\right)k^2 $$ therefore $$ \Re f(k,\theta) = \frac{-\frac{1}{a} + \frac{1}{2}r_ek^2}{\frac{1}{a^2} +\left(1 -\frac{r_e}{a}\right)k^2} = \frac{-a +\frac{a^2}{2}r_ek^2}{1 +a^2\left(1 -\frac{r_e}{a}\right)k^2} $$ using $$ \frac{1}{1+x} = 1 - x +... $$ we obtain $$ \Re f(k,\theta) = \left(-a +\frac{a^2}{2}r_ek^2\right)\left(1 - a^2\left(1 -\frac{r_e}{a}\right)k^2 + ..O(k^4)\right)\\ = -a +a^3\left(1 -\frac{r_e}{a}\right)k^2 + \frac{a^2}{2}r_ek^2\\ =-a + a^3k^2 -a^2r_ek^2 + \frac{a^2}{2}r_ek^2 = -a +a^3k^2 -\frac{a^2}{2}r_ek^2 $$ This yields the result as required $$ \Re f(k,\theta) = -a +a^2\left(a - \frac{r_e}{2}\right)k^2. $$