Fdtxjr

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

$$z_pm:=frac12 left(zpm sqrtz^2-4 right).$$

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 $mathcalR tomathcalC$ therefore (I THINK, HAVE NOT BEEN ABLE TO FIND PROOF YET) the Taylor expansion of $textIm(z_pm(epsilon))$ 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

$fracdzdepsilon|_epsilon = 0 = -irexp(ipi)exp(-iepsilon)|_epsilon = 0 = ir$

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:

$$z_pm|_epsilon = 0 =frac12left(a -r pm sqrt(a-r)^2 -4right) = frac12left(a -r pm sqrta -r -2sqrta-r+2right).$$

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:
$$fracdz_pmdepsilon|_epsilon = 0 = frac12left(fracdzdepsilon pm fracdzdepsilon fraczsqrtz^2 - 4right) = fracir2left(1 pm fraca-rsqrt(a-r)^2 -4right)$$

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:
$$z_pm(epsilon) approx frac12left(a -r pm sqrt(a-r)^2 -4right) +fracir2left(1 pm fraca-rsqrt(a-r)^2 -4right) epsilon$$

In case $a -2 le r le a+2$ we get:
$$textImleft(z_pm(epsilon)right) approx pm frac12left(|sqrta -r -2|sqrta-r+2right) + fracr2epsilon$$

and in case $r le a-2$:

$$textImleft(z_pm(epsilon)right) approx fracr2left(1 pm fraca-rsqrt(a-r)^2 -4right) epsilon$$

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.

$textbfFull Edition.$

$colorbrowntextbfExact expression of the imaginary part.$

Denote
$$z_1 = y = u+iv,$$
then $y_1+y_2 = z = a-re^-ivarepsilon,quad y_1y_2=1,quad$ so
$$y^2-zy+1=0tag1,$$
$$(u+iv)^2-(a-rcosvarepsilon+irsinvarepsilon)(u+iv)+1=0,$$
begincases
u^2-v^2 - u(a-rcosvarepsilon) + vrsinvarepsilon +1 = 0\
2uv - ursinvarepsilon - v(a-rcosvarepsilon) =0,
endcases
begincases
u=dfrac v(a-rcosvarepsilon)2v-rsinvarepsilon\[4pt]
left(dfrac (a-rcosvarepsilon)^2(2v-rsinvarepsilon)^2-1right)v^2
- left(dfrac (a-rcosvarepsilon)^22v-rsinvarepsilon - rsinvarepsilonright)v +1 = 0,\
endcases
$$(a- rcosvarepsilon)^2(v^2-v(2v-rsinvarepsilon)) - (v^2-vrsinvarepsilon-1)(2v-rsinvarepsilon)^2 = 0,$$
or
$$f(t,varepsilon) = t^4+((a-rcosvarepsilon)^2-r^2sin^2varepsilon-4)t^2-(a-rcosvarepsilon)^2r^2sin^2varepsilon=0,tag2.1$$
where
$$t=2v-rsinvarepsilon.tag2.2$$

Discriminant of the biquadratic equation $(2.1)$ is
$$D=((a-rcosvarepsilon-2)^2+r^2sin^2varepsilon)((a-rcosvarepsilon+2)^2+r^2sin^2varepsilon) > 0,tag3.1$$
and the explicit expression for $v(varepsilon)$ is
$$v(varepsilon)=dfracrsinvarepsilonpm t(varepsilon)2,quadtextwherequad t(varepsilon) = sqrtdfracr^2sin^2varepsilon+4-(a-rcosvarepsilon)^2 +sqrt D2.tag3.2$$

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.

$colorbrowntextbfData transformations.$

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
$$g(t,p)=t^4+b(p)t^2+c(p),tag5.1$$
beginalign
&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
endalign
$$begincases
b(p)=2r^2p^2+2r(a-2r)p+(a-r)^2-4\
c(p) = r^4p^4+2r^3(a-2r)p^3+r^2(a^2-6ar+5r^2)p^2-2r^2(a-r)^2p
endcasestag5.2$$

Will be built Maclaurin series $t(p)$ with
$$t(0)= t_0 = sqrt4-(a-r)^2phantombig.tag6$$

Required derivatives will be obtained through differentiation of $g(t,p),$
i.e. the expression $(5).$

$colorbrowntextbfImplicit differentiation.$

Denote
$$g_ij(t,p)=dfracpartial^(i+j)gpartial t^i partial p^j,tag7.1$$

Taking in account that

$$beginalign &dfrac ddpvarphi(t,p) = varphi,'_t,t'+varphi'_p,\[4pt]
&dfrac ddpvarphi(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,endalign$$

repeated re-differentiation of $(5.1)$ becames simple:
$$beginaligned
&dfrac dgdp = g_10t'+g_01,\[4pt]
&dfrac d^2gdp^2 = dfrac ddp(g_10t'+g_01)
= Bigl(g_10t''+g_20t'^,2,+g_11t'Bigr)+g_11t'+ g_02\[4pt]
&= g_10t''+g_20t'^,2+2g_11t'+ g_02,\[4pt]
&dfrac d^3gdp^3
= dfrac ddpBigl(g_10t''+g_20t'^,2+2g_11t'+g_02Bigr)\[4pt]
&= (g_10t'''+g_20t't''+g_11t'')
+(2g_20t't''+g_30t'^,3+g_21t'^,2)\[4pt]
&+2(g_11t''+g_21t'^,2+g_12t')
+(g_12t'+g_03)\[4pt]
&= g_10t''' +3g_20t't'' +3g_11t''
+g_30t'^,3+3g_21t'^,2+3g_12t'+g_03,\[4pt]
&dfrac d^4gdp^4
= dfrac ddpBigl(g_10t''' +3g_20t't'' + 3g_11t''
+g_30t'^,3+3g_21t'^,2+3g_12t'+g_03Bigr)\[4pt]
&= (g_10t^IV+g_20t't'''+g_11t''')
+3(g_20(t't'''+t''^,2)+g_30t'^,2t''+g_21t't'')\[4pt]
&+3(g_11t'''+g_21t't''+g_12t'')
+(3g_30t'^,2t''+g_40t'^,4+g_31t'^,3)\[4pt]
&+3(2g_21t't'' +g_31t'^,3+g_22t'^,2)
+3(g_12t''+g_22t'^,2+g_13t')+(g_13t'+g_04)\[4pt]
&= g_10t^IV+g_20(4t't'''+3t''^,2)+4g_11t'''
+6g_30t'^,2t''+12g_21t't''+6g_12t''\[4pt]
&+g_40t'^,4+4g_31t'^,3
+6g_22t'^,2+4g_13t'+g_04dots
endalignedtag7.2$$

Partial derivatives in the point $(t_0,0)$ can be calculated by formula
$$g_ij= delta_j0T_4i+B_jT_2i+delta_i0C_j,tag8.1$$
where $delta_ij$ is Kronecker symbol,
$$beginpmatrixT_40 \ T_41 \ T_42 \ T_43 \ T_44endpmatrix
=beginpmatrixt_0^4 \4t_0^3 \12t_0^2 \24t_0 \24 endpmatrix,quad
beginpmatrixT_20 \ T_21 \ T_22 \ T_23 \ T_24endpmatrix
=beginpmatrixt_0^2 \2t_0 \2 \0 \0 endpmatrix,tag8.2$$

$$beginpmatrixB_0 \ B_1 \ B_2 \ B_3 \ B_4endpmatrix
=beginpmatrix-t_0^2 \2ar-4r^2 \4r^2 \0 \0 endpmatrix,quad
beginpmatrixC_0 \ C_1 \ C_2 \ C_3 \ C_4endpmatrix
=beginpmatrix0 \-2r^2(a-r)^2 \ 2r^2(a-r)(a-5r) \12r^3(a-2r) \24r^4 endpmatrix.tag8.3$$

The derivatives' values $g_ij(t_0,0)$ can be presented in the table form of
$$left[beginmatrix
g_ij(t_0,0) & j=0 & j=1 & j=2 & j=3 & j=4 \
i=0 & 0 & 2r(a-2r)t_0^2-2r^2(a-r)^2 & 4r^2t_0^2 +2r^2(a-r)(a-5r)& 12r^3(a-2r) & 24r^4 \
i=1 & 2t_0^3 & 4r(a-2r)t_0 & 8r^2t_0 & 0 && \
i=2 & 10t_0^2 & 4r(a-2r) & 8r^2 &&& \
i=3 & 24t_0 & 0 &&&& \
i=4 & 24 & &&&& \
endmatrixright]tag9$$
All the other derivatives' values $g_ij(t_0,0)$ equal to zero.

$colorbrowntextbfDerivatives for the series.$

Obtained results $(7),(9)$ allow to write the simple system for $t'(0),t''(0),t'''(0),t^IV(0)$ in the form of

$$beginaligned
&dfrac dgdp(t_0,0) = bigg(g_10t'+g_01bigg)bigg|_(t_0,0) = 0,\[4pt]
&dfrac d^2gdp^2(t_0,0)= bigg(g_10t''+g_20t'^,2,+2g_11t'+g_02bigg)bigg|_(t_0,0)=0,\[4pt]
&dfrac d^3gdp^3(t_0,0) = bigg(g_10t''' +3(g_20t'+g_11)t''
+g_30t'^,3+3g_21t'^,2+3g_12t'+g_03bigg)bigg|_(t_0,0)=0,\[4pt]
&dfrac d^4gdp^4(t_0,0) = bigg(g_10t^IV+4(g_20t'+g_11)t''' +3g_20t''^,2+6(g_30t'^,2+2g_21t'+g_12)t''\[4pt]
&+g_40t'^,4+4g_31t'^,3+6g_22t'^,2+4g_13t'+g_04bigg)bigg|_(t_0,0)=0dots.
endalignedtag10$$

The system $(10)$ allows to get the explicit expressions for the required derivatives
$$tau_1=t'(0), tau_2=t''(0), tau_3=t'''(0), tau_4=t^IV(0). tag11$$

There are
$$beginaligned
&tau_1 = - dfracg_01g_10,\[4pt]
&tau_2 = - dfrac1g_10 bigg(g_20tau_1^2+2g_11tau_1+g_02bigg),\[4pt]
&tau_3 = - dfrac1g_10 bigg(3(g_20tau_1+g_11)tau_2
+g_30tau_1^3+3g_21tau_1^2+3g_12tau_1+g_03bigg),\[4pt]
&tau_4 = - dfrac1g_10 bigg(4(g_20tau_1+g_11)tau_3 +3g_20tau_2^2+6(g_30tau_1^2+2g_21tau_1+g_12)tau_2
+g_40tau_1^4+6g_22tau_1^2+g_04bigg),\[4pt]
&dots,
endalignedtag12$$
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.

$colorbrowntextbfMaclaurin series of 9th order.$

Obtained series has the form of
$$beginalign
&t(varepsilon) = t_0 + tau_1(1-cosvarepsilon)+frac12tau_2(1-cosvarepsilon)^2+frac16tau_3(1-cosvarepsilon)^3+frac124tau_4(1-cosvarepsilon)^4+dots\[4pt]
&= t_0 + tau_1left(frac12!varepsilon^2 -frac14!varepsilon^4 +frac16!varepsilon^6-frac18!varepsilon^8right)
+frac12tau_2left(frac12!varepsilon^2-frac14!varepsilon^4 +frac16!varepsilon^6right)^2\[4pt]
&+frac16tau_3left(frac12!varepsilon^2-frac14!varepsilon^4right)^3 +frac124tau_4left(frac12!varepsilon^2right)^4+dots
= t_0 + frac12tau_1varepsilon^2 \[4pt]
&+frac124(-tau_1+6tau_2)varepsilon^4+frac1720(tau_1-15tau_2+15tau_3)varepsilon^6
+frac140320(-tau_1+63tau_2-210tau_3+105tau_4+dots
endalign$$

Then, in accordance with $(3)-(4),$ can be obtained Maclaurin series for the both branches in the form of
$$beginalign
&v(varepsilon)=frac 12(rsinvarepsilon pm t(varepsilon))
= pmfrac12 t_0 + frac r2 varepsilon pm frac14 tau_1varepsilon^2 -frac r12 varepsilon^3 pm frac148(-tau_1+6tau_1)varepsilon^4 +frac r240varepsilon^5\[4pt]
& pm frac11440(tau_1-15tau_2+15tau_3)varepsilon^6
- frac r10080varepsilon^7
pm frac180640(-tau_1+63tau_2-210tau_3+105tau_4)varepsilon^8\[4pt]
&+ frac r362880varepsilon^9+dots
endalign$$

Therefore, Maclaurin series of nineth order $colorgreentextrmcan be built$.

Easy to see that used approach allows to calculate the arbitrary quantity of the series terms.