Taylor expansion of imaginary part?-Doable or not? I have a number $z = a+re^{i(\pi-\varepsilon)}$ and $\varepsilon>0$ is small, $a,r>0.$
You can assume furthermore that $r\le a+2.$
I then define the expressions 
$$z_{\pm}:=\frac{1}{2} \left(z\pm \sqrt{z^2-4} \right).$$
The question is: Can one find a Taylor expansion of the imaginary part of $z_{\pm}$ in terms of $\varepsilon$. I would like to know at least what the leading order terms are for $\varepsilon$ small. 
Let me finish with a quote of encouragement:
Mark Twain — 'They did not know it was impossible so they did it'
 A: My approach is based on the fact that $z_{\pm}(\epsilon)$ is a function from $\mathcal{R} \to\mathcal{C}$ therefore (I THINK, HAVE NOT BEEN ABLE TO FIND PROOF YET) the Taylor expansion of $\text{Im}(z_{\pm}(\epsilon))$ is the same as Imaginary part of the Taylor expansion of $z_{\pm}(\epsilon)$.
Firstly write:
$ z = a + r \exp(i\pi)\exp(-i\epsilon)$.
Then realize: $z(\epsilon = 0 ) = a -r$ and
$\frac{dz}{d\epsilon}|_{\epsilon = 0} = -ir\exp(i\pi)\exp(-i\epsilon)|_{\epsilon = 0} = ir$
We will now compute the first two terms of the Taylor expansion of $z_{\pm}$ in terms around $\epsilon = 0$. We start at zeroth order:
$$z_{\pm}|_{\epsilon = 0} =\frac{1}{2}\left(a -r \pm \sqrt{(a-r)^2 -4}\right) = \frac{1}{2}\left(a -r \pm \sqrt{a -r -2}\sqrt{a-r+2}\right).$$
Note here that since $r \le a + 2$ the second square root is always positive. If also $r \le a -2$ then the first square root is positive and the whole term will be reall. In case $a -2 \le r \le a+2$ we will have a square root of a negative number and thus we will get Imaginary numbers from there. Thus, the leading order term of the expansion will be $\epsilon^0$.
And now do the first order term:
$$\frac{dz_{\pm}}{d\epsilon}|_{\epsilon = 0} = \frac{1}{2}\left(\frac{dz}{d\epsilon} \pm \frac{dz}{d\epsilon} \frac{z}{\sqrt{z^2 - 4}}\right) = \frac{ir}{2}\left(1 \pm \frac{a-r}{\sqrt{(a-r)^2 -4}}\right)$$
The same distinction needs to be made here if $a -2 \le r \le a+2$ the square root will give us imaginary numbers.
Putting it all together we get: 
$$z_{\pm}(\epsilon) \approx \frac{1}{2}\left(a -r \pm \sqrt{(a-r)^2 -4}\right) +\frac{ir}{2}\left(1 \pm \frac{a-r}{\sqrt{(a-r)^2 -4}}\right) \epsilon$$
In case $a -2 \le r \le a+2$ we get:
$$\text{Im}\left(z_{\pm}(\epsilon)\right) \approx \pm \frac{1}{2}\left(|\sqrt{a -r -2}|\sqrt{a-r+2}\right) + \frac{r}{2}\epsilon$$
and in case $r \le a-2$: 
$$\text{Im}\left(z_{\pm}(\epsilon)\right) \approx \frac{r}{2}\left(1 \pm \frac{a-r}{\sqrt{(a-r)^2 -4}}\right) \epsilon$$
Edit: Getting the full expansion shouldn't be too hard once you are aware you can expand $z_{\pm}$ and then take the imaginary part.
