TITLE: Taylor expansion of imaginary part?-Doable or not?
QUESTION [7 upvotes]: 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'

REPLY [3 votes]: $\textbf{Full Edition.}$
$\color{brown}{\textbf{Exact expression of the imaginary part.}}$
Denote
$$z_1 = y = u+iv,$$
then $y_1+y_2 = z = a-re^{-i\varepsilon},\quad y_1y_2=1,\quad$ so
$$y^2-zy+1=0\tag1,$$
$$(u+iv)^2-(a-r\cos\varepsilon+ir\sin\varepsilon)(u+iv)+1=0,$$
\begin{cases}
u^2-v^2 - u(a-r\cos\varepsilon) + vr\sin\varepsilon  +1 = 0\\
2uv - ur\sin\varepsilon  - v(a-r\cos\varepsilon) =0,
\end{cases}
\begin{cases}
u=\dfrac {v(a-r\cos\varepsilon)}{2v-r\sin\varepsilon}\\[4pt]
\left(\dfrac {(a-r\cos\varepsilon)^2}{(2v-r\sin\varepsilon)^2}-1\right)v^2 
- \left(\dfrac {(a-r\cos\varepsilon)^2}{2v-r\sin\varepsilon} - r\sin\varepsilon\right)v  +1 = 0,\\
\end{cases}
$$(a- r\cos\varepsilon)^2(v^2-v(2v-r\sin\varepsilon)) - (v^2-vr\sin\varepsilon-1)(2v-r\sin\varepsilon)^2 = 0,$$
or
$$f(t,\varepsilon) = t^4+((a-r\cos\varepsilon)^2-r^2\sin^2\varepsilon-4)t^2-(a-r\cos\varepsilon)^2r^2\sin^2\varepsilon=0,\tag{2.1}$$
where 
$$t=2v-r\sin\varepsilon.\tag{2.2}$$
Discriminant of the biquadratic equation $(2.1)$ is
$$D=((a-r\cos\varepsilon-2)^2+r^2\sin^2\varepsilon)((a-r\cos\varepsilon+2)^2+r^2\sin^2\varepsilon) > 0,\tag{3.1}$$
and the explicit expression for $v(\varepsilon)$ is
$$v(\varepsilon)=\dfrac{r\sin\varepsilon\pm t(\varepsilon)}2,\quad\text{where}\quad t(\varepsilon) = \sqrt{\dfrac{r^2\sin^2\varepsilon+4-(a-r\cos\varepsilon)^2 +\sqrt D}2}.\tag{3.2}$$
From $(2.1)$ follows that
$$(t^2)_1(t^2)_2 = -(a-r\cos\varepsilon)^2r^2\sin^2\varepsilon.$$ 
I.e. real Taylor series for the other solutions cannot be built.
$\color{brown}{\textbf{Data transformations.}}$
Explicit expression $(3)$ of the imaginary part looks too hard for repeated re-differentiation. 
At the same time, the equation $(2)$ has linear structure and allows suitable repeated re-differentiation. This way required some additional operations.
Firstly, let us present $(2)$ via superposition in the form of
$$f(t,\varepsilon) = g(t, 1-\cos\varepsilon)),\tag4$$
where
$$g(t,p)=t^4+b(p)t^2+c(p),\tag{5.1}$$
\begin{align}
&b(p)=(a-r(1-p))^2-r^2(1-(1-p)^2)-4 = 2r^2p^2+2r(a-2r)p+((a-r)^2-4)
,\\[4pt]
&c(p)= -r^2(a-r(1-p))^2(1-(1-p)^2)\\[4pt]
& =  r^4p^4+2r^3(a-2r)p^3+r^2(a^2-6ar+5r^2)p^2-2r^2(a-r)^2p
\end{align}
$$\begin{cases}
b(p)=2r^2p^2+2r(a-2r)p+(a-r)^2-4\\
c(p) = r^4p^4+2r^3(a-2r)p^3+r^2(a^2-6ar+5r^2)p^2-2r^2(a-r)^2p
\end{cases}\tag{5.2}$$
Will be built Maclaurin series $t(p)$  with 
$$t(0)= t_0 = \sqrt{4-(a-r)^2\phantom{\big|}}.\tag6$$
Required derivatives will be obtained through differentiation of $g(t,p),$
i.e. the expression $(5).$
$\color{brown}{\textbf{Implicit differentiation.}}$
Denote
$$g_{ij}(t,p)=\dfrac{\partial^{(i+j)}g}{\partial t^i \partial p^j},\tag{7.1}$$
Taking in account that

$$\begin{align} &\dfrac d{dp}\varphi(t,p) = \varphi\,'_t\,t'+\varphi'_p,\\[4pt] 
 &\dfrac d{dp}\varphi(t,p)\psi(t^{(n)},\dots,t'',t') 
 = \varphi\psi\,'_{t^{(n)}}\,t^{(n+1)} +\dots+\varphi\psi\,'_{t''}\,t''' +\varphi\psi\,'_{t'}\,t''+\varphi'_t\psi\,t'+\varphi'_p\psi,\end{align}$$

repeated re-differentiation of $(5.1)$ becames simple:
$$\begin{aligned}
&\dfrac {dg}{dp} = g_{10}t'+g_{01},\\[4pt]
&\dfrac {d^2g}{dp^2} = \dfrac d{dp}(g_{10}t'+g_{01})
= \Bigl(g_{10}t''+g_{20}t'^{\,2}\,+g_{11}t'\Bigr)+g_{11}t'+ g_{02}\\[4pt]
&= g_{10}t''+g_{20}t'^{\,2}+2g_{11}t'+ g_{02},\\[4pt]
&\dfrac {d^3g}{dp^3} 
= \dfrac d{dp}\Bigl(g_{10}t''+g_{20}t'^{\,2}+2g_{11}t'+g_{02}\Bigr)\\[4pt]
&= (g_{10}t'''+g_{20}t't''+g_{11}t'')
+(2g_{20}t't''+g_{30}t'^{\,3}+g_{21}t'^{\,2})\\[4pt]
&+2(g_{11}t''+g_{21}t'^{\,2}+g_{12}t')
+(g_{12}t'+g_{03})\\[4pt]
&= g_{10}t''' +3g_{20}t't'' +3g_{11}t''
+g_{30}t'^{\,3}+3g_{21}t'^{\,2}+3g_{12}t'+g_{03},\\[4pt]
&\dfrac {d^4g}{dp^4} 
= \dfrac d{dp}\Bigl(g_{10}t''' +3g_{20}t't'' + 3g_{11}t''
+g_{30}t'^{\,3}+3g_{21}t'^{\,2}+3g_{12}t'+g_{03}\Bigr)\\[4pt]
&= (g_{10}t^{IV}+g_{20}t't'''+g_{11}t''')
+3(g_{20}(t't'''+t''^{\,2})+g_{30}t'^{\,2}t''+g_{21}t't'')\\[4pt]
&+3(g_{11}t'''+g_{21}t't''+g_{12}t'')
+(3g_{30}t'^{\,2}t''+g_{40}t'^{\,4}+g_{31}t'^{\,3})\\[4pt]
&+3(2g_{21}t't'' +g_{31}t'^{\,3}+g_{22}t'^{\,2})
+3(g_{12}t''+g_{22}t'^{\,2}+g_{13}t')+(g_{13}t'+g_{04})\\[4pt]
&= g_{10}t^{IV}+g_{20}(4t't'''+3t''^{\,2})+4g_{11}t'''
+6g_{30}t'^{\,2}t''+12g_{21}t't''+6g_{12}t''\\[4pt]
&+g_{40}t'^{\,4}+4g_{31}t'^{\,3}
+6g_{22}t'^{\,2}+4g_{13}t'+g_{04}\dots
\end{aligned}\tag{7.2}$$
Partial derivatives in the point $(t_0,0)$ can be calculated by formula
$$g_{ij}= \delta_{j0}T_{4i}+B_jT_{2i}+\delta_{i0}C_j,\tag{8.1}$$
where $\delta_{ij}$ is Kronecker symbol,
$$\begin{pmatrix}T_{40} \\ T_{41} \\ T_{42} \\ T_{43} \\ T_{44}\end{pmatrix}
=\begin{pmatrix}t_0^4 \\4t_0^3 \\12t_0^2 \\24t_0 \\24 \end{pmatrix},\quad
\begin{pmatrix}T_{20} \\ T_{21} \\ T_{22} \\ T_{23} \\ T_{24}\end{pmatrix}
=\begin{pmatrix}t_0^2 \\2t_0 \\2 \\0 \\0 \end{pmatrix},\tag{8.2}$$
$$\begin{pmatrix}B_0 \\ B_1 \\ B_2 \\ B_3 \\ B_4\end{pmatrix}
=\begin{pmatrix}-t_0^2 \\2ar-4r^2 \\4r^2 \\0 \\0 \end{pmatrix},\quad
\begin{pmatrix}C_0 \\ C_1 \\ C_2 \\ C_3 \\ C_4\end{pmatrix}
=\begin{pmatrix}0 \\-2r^2(a-r)^2 \\ 2r^2(a-r)(a-5r) \\12r^3(a-2r) \\24r^4 \end{pmatrix}.\tag{8.3}$$
The derivatives' values $g_{ij}(t_0,0)$ can be presented in the table form of
$$\left[\begin{matrix}
g_{ij}(t_0,0) & j=0 & j=1 & j=2 & j=3 & j=4 \\
i=0 &  0 & 2r(a-2r)t_0^2-2r^2(a-r)^2 & 4r^2t_0^2 +2r^2(a-r)(a-5r)& 12r^3(a-2r) & 24r^4 \\
i=1 & 2t_0^3 & 4r(a-2r)t_0 & 8r^2t_0 & 0 && \\
i=2 & 10t_0^2 & 4r(a-2r) & 8r^2 &&& \\
i=3 & 24t_0 & 0 &&&& \\
i=4 & 24 &  &&&& \\
\end{matrix}\right]\tag9$$
All the other derivatives' values $g_{ij}(t_0,0)$ equal to zero.
$\color{brown}{\textbf{Derivatives for the series.}}$
Obtained results $(7),(9)$ allow to write the simple system for $t'(0),t''(0),t'''(0),t^{IV}(0)$ in the form of
$$\begin{aligned}
&\dfrac {dg}{dp}(t_0,0) = \bigg(g_{10}t'+g_{01}\bigg)\bigg|_{(t_0,0)} = 0,\\[4pt]
&\dfrac {d^2g}{dp^2}(t_0,0)= \bigg(g_{10}t''+g_{20}t'^{\,2}\,+2g_{11}t'+g_{02}\bigg)\bigg|_{(t_0,0)}=0,\\[4pt]
&\dfrac {d^3g}{dp^3}(t_0,0) = \bigg(g_{10}t''' +3(g_{20}t'+g_{11})t'' 
+g_{30}t'^{\,3}+3g_{21}t'^{\,2}+3g_{12}t'+g_{03}\bigg)\bigg|_{(t_0,0)}=0,\\[4pt]
&\dfrac {d^4g}{dp^4}(t_0,0) = \bigg(g_{10}t^{IV}+4(g_{20}t'+g_{11})t''' +3g_{20}t''^{\,2}+6(g_{30}t'^{\,2}+2g_{21}t'+g_{12})t''\\[4pt]
&+g_{40}t'^{\,4}+4g_{31}t'^{\,3}+6g_{22}t'^{\,2}+4g_{13}t'+g_{04}\bigg)\bigg|_{(t_0,0)}=0\dots.
\end{aligned}\tag{10}$$
The system $(10)$ allows to get the explicit expressions for the required derivatives
$$\tau_1=t'(0),\  \tau_2=t''(0),\  \tau_3=t'''(0),\  \tau_4=t^{IV}(0). \tag{11}$$ 
There are
$$\begin{aligned}
&\tau_1 = - \dfrac{g_{01}}{g_{10}},\\[4pt]
&\tau_2 = - \dfrac{1}{g_{10}} \bigg(g_{20}\tau_1^2+2g_{11}\tau_1+g_{02}\bigg),\\[4pt]
&\tau_3 = - \dfrac{1}{g_{10}} \bigg(3(g_{20}\tau_1+g_{11})\tau_2 
+g_{30}\tau_1^3+3g_{21}\tau_1^2+3g_{12}\tau_1+g_{03}\bigg),\\[4pt]
&\tau_4 = - \dfrac{1}{g_{10}} \bigg(4(g_{20}\tau_1+g_{11})\tau_3 +3g_{20}\tau_2^2+6(g_{30}\tau_1^2+2g_{21}\tau_1+g_{12})\tau_2
+g_{40}\tau_1^4+6g_{22}\tau_1^2+g_{04}\bigg),\\[4pt]
&\dots,
\end{aligned}\tag{12}$$
wherein all unzero values $g_{ij}$ are defined in the table $(9).$
Expressions for the next $\tau_k$ can not contain essentially greater quantity of terms, because all the next partial derivatives are zeros. 
$\color{brown}{\textbf{Maclaurin series of 9th order.}}$
Obtained series has the form of
$$\begin{align}
&t(\varepsilon) = t_0 + \tau_1(1-\cos\varepsilon)+\frac12\tau_2(1-\cos\varepsilon)^2+\frac16\tau_3(1-\cos\varepsilon)^3+\frac1{24}\tau_4(1-\cos\varepsilon)^4+\dots\\[4pt]
&= t_0 + \tau_1\left(\frac1{2!}\varepsilon^2 -\frac1{4!}\varepsilon^4 +\frac1{6!}\varepsilon^6-\frac1{8!}\varepsilon^8\right)
+\frac12\tau_2\left(\frac1{2!}\varepsilon^2-\frac1{4!}\varepsilon^4 +\frac1{6!}\varepsilon^6\right)^2\\[4pt]
&+\frac16\tau_3\left(\frac1{2!}\varepsilon^2-\frac1{4!}\varepsilon^4\right)^3 +\frac1{24}\tau_4\left(\frac1{2!}\varepsilon^2\right)^4+\dots
= t_0 + \frac12\tau_1\varepsilon^2 \\[4pt]
&+\frac1{24}(-\tau_1+6\tau_2)\varepsilon^4+\frac1{720}(\tau_1-15\tau_2+15\tau_3)\varepsilon^6 
+\frac1{40320}(-\tau_1+63\tau_2-210\tau_3+105\tau_4+\dots
\end{align}$$
Then, in accordance with $(3)-(4),$ can be obtained Maclaurin series for the both branches in the form of
$$\begin{align}
&v(\varepsilon)=\frac 12(r\sin\varepsilon \pm t(\varepsilon))
= \pm\frac12 t_0 + \frac r2 \varepsilon \pm \frac14 \tau_1\varepsilon^2 -\frac r{12} \varepsilon^3 \pm \frac1{48}(-\tau_1+6\tau_1)\varepsilon^4 +\frac r{240}\varepsilon^5\\[4pt]
& \pm \frac1{1440}(\tau_1-15\tau_2+15\tau_3)\varepsilon^6 
- \frac r{10080}\varepsilon^7 
\pm \frac1{80640}(-\tau_1+63\tau_2-210\tau_3+105\tau_4)\varepsilon^8\\[4pt] 
&+ \frac r{362880}\varepsilon^9+\dots
\end{align}$$
Therefore, Maclaurin series of nineth order $\color{green}{\textrm{can be built}}$. 
Easy to see that used approach allows to calculate the arbitrary quantity of the series terms.