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I have a number $z = a+re^{i(pi-varepsilon)}$ and $varepsilon>0$ is small, $a,r>0.$

You can assume furthermore that $rle a+2.$

I then define the expressions

$$z_{pm}:=frac{1}{2} left(zpm sqrt{z^2-4} right).$$

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'

My approach is based on the fact that $z_{pm}(epsilon)$ is a function from $mathcal{R} tomathcal{C}$ 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(ipi)exp(-iepsilon)$.

Then realize: $z(epsilon = 0 ) = a -r$ and

$frac{dz}{depsilon}|_{epsilon = 0} = -irexp(ipi)exp(-iepsilon)|_{epsilon = 0} = ir$

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:
$$frac{dz_{pm}}{depsilon}|_{epsilon = 0} = frac{1}{2}left(frac{dz}{depsilon} pm frac{dz}{depsilon} frac{z}{sqrt{z^2 - 4}}right) = frac{ir}{2}left(1 pm frac{a-r}{sqrt{(a-r)^2 -4}}right)$$

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.

$textbf{Full Edition.}$

$color{brown}{textbf{Exact expression of the imaginary part.}}$

Denote
$$z_1 = y = u+iv,$$
then $y_1+y_2 = z = a-re^{-ivarepsilon},quad y_1y_2=1,quad$ so
$$y^2-zy+1=0tag1,$$
$$(u+iv)^2-(a-rcosvarepsilon+irsinvarepsilon)(u+iv)+1=0,$$
begin{cases}
u^2-v^2 - u(a-rcosvarepsilon) + vrsinvarepsilon +1 = 0\
2uv - ursinvarepsilon - v(a-rcosvarepsilon) =0,
end{cases}
begin{cases}
u=dfrac {v(a-rcosvarepsilon)}{2v-rsinvarepsilon}\[4pt]
left(dfrac {(a-rcosvarepsilon)^2}{(2v-rsinvarepsilon)^2}-1right)v^2
- left(dfrac {(a-rcosvarepsilon)^2}{2v-rsinvarepsilon} - rsinvarepsilonright)v +1 = 0,\
end{cases}
$$(a- rcosvarepsilon)^2(v^2-v(2v-rsinvarepsilon)) - (v^2-vrsinvarepsilon-1)(2v-rsinvarepsilon)^2 = 0,$$
or
$$f(t,varepsilon) = t^4+((a-rcosvarepsilon)^2-r^2sin^2varepsilon-4)t^2-(a-rcosvarepsilon)^2r^2sin^2varepsilon=0,tag{2.1}$$
where
$$t=2v-rsinvarepsilon.tag{2.2}$$

Discriminant of the biquadratic equation $(2.1)$ is
$$D=((a-rcosvarepsilon-2)^2+r^2sin^2varepsilon)((a-rcosvarepsilon+2)^2+r^2sin^2varepsilon) > 0,tag{3.1}$$
and the explicit expression for $v(varepsilon)$ is
$$v(varepsilon)=dfrac{rsinvarepsilonpm t(varepsilon)}2,quadtext{where}quad t(varepsilon) = sqrt{dfrac{r^2sin^2varepsilon+4-(a-rcosvarepsilon)^2 +sqrt D}2}.tag{3.2}$$

From $(2.1)$ follows that
$$(t^2)_1(t^2)_2 = -(a-rcosvarepsilon)^2r^2sin^2varepsilon.$$
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
$$f(t,varepsilon) = g(t, 1-cosvarepsilon)),tag4$$
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
$$t(0)= t_0 = sqrt{4-(a-r)^2phantom{big|}}.tag6$$

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')
= varphipsi,'_{t^{(n)}},t^{(n+1)} +dots+varphipsi,'_{t''},t''' +varphipsi,'_{t'},t''+varphi'_tpsi,t'+varphi'_ppsi,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,
$$begin{pmatrix}T_{40} \ T_{41} \ T_{42} \ T_{43} \ T_{44}end{pmatrix}
=begin{pmatrix}t_0^4 \4t_0^3 \12t_0^2 \24t_0 \24 end{pmatrix},quad
begin{pmatrix}T_{20} \ T_{21} \ T_{22} \ T_{23} \ T_{24}end{pmatrix}
=begin{pmatrix}t_0^2 \2t_0 \2 \0 \0 end{pmatrix},tag{8.2}$$

$$begin{pmatrix}B_0 \ B_1 \ B_2 \ B_3 \ B_4end{pmatrix}
=begin{pmatrix}-t_0^2 \2ar-4r^2 \4r^2 \0 \0 end{pmatrix},quad
begin{pmatrix}C_0 \ C_1 \ C_2 \ C_3 \ C_4end{pmatrix}
=begin{pmatrix}0 \-2r^2(a-r)^2 \ 2r^2(a-r)(a-5r) \12r^3(a-2r) \24r^4 end{pmatrix}.tag{8.3}$$

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)}=0dots.
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-cosvarepsilon)+frac12tau_2(1-cosvarepsilon)^2+frac16tau_3(1-cosvarepsilon)^3+frac1{24}tau_4(1-cosvarepsilon)^4+dots\[4pt]
&= t_0 + tau_1left(frac1{2!}varepsilon^2 -frac1{4!}varepsilon^4 +frac1{6!}varepsilon^6-frac1{8!}varepsilon^8right)
+frac12tau_2left(frac1{2!}varepsilon^2-frac1{4!}varepsilon^4 +frac1{6!}varepsilon^6right)^2\[4pt]
&+frac16tau_3left(frac1{2!}varepsilon^2-frac1{4!}varepsilon^4right)^3 +frac1{24}tau_4left(frac1{2!}varepsilon^2right)^4+dots
= t_0 + frac12tau_1varepsilon^2 \[4pt]
&+frac1{24}(-tau_1+6tau_2)varepsilon^4+frac1{720}(tau_1-15tau_2+15tau_3)varepsilon^6
+frac1{40320}(-tau_1+63tau_2-210tau_3+105tau_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(rsinvarepsilon pm t(varepsilon))
= pmfrac12 t_0 + frac r2 varepsilon pm frac14 tau_1varepsilon^2 -frac r{12} varepsilon^3 pm frac1{48}(-tau_1+6tau_1)varepsilon^4 +frac r{240}varepsilon^5\[4pt]
& pm frac1{1440}(tau_1-15tau_2+15tau_3)varepsilon^6
- frac r{10080}varepsilon^7
pm frac1{80640}(-tau_1+63tau_2-210tau_3+105tau_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.