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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 $r\le a+2.$

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'

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2 Answers 2

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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.

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  • $\begingroup$ The reason why mapping from $R \to C$ should obey the Taylor relation I mentioned is because if both $f(x)$ and $\text{Im}(f(x))$ are analytical (meaning the expansion exists) then we must have that the series converge to each other if we look at the Imaginary part due to uniqueness (the same holds for the real part). For functions $C \to C$ this does not hold! Say $f(z) =(2+i) z$ then we can't say $Im(f(z)) = Im(2+i) \times Im(z)$ because $z$ is an complex number! $\endgroup$ Mar 6, 2019 at 11:22
  • $\begingroup$ Your result seems to imply that if $r=a$ the $Im(z_{\pm}(\varepsilon)) \approx \frac{r}{2} \varepsilon. $ Now, take $r=a$, then I think it becomes apparent that this result cannot be true. Just look at what $z_{\pm}$ is for $\varepsilon=0.$ $\endgroup$
    – user505183
    Mar 6, 2019 at 11:25
  • $\begingroup$ Indeed, I require a>r+2. $\endgroup$ Mar 6, 2019 at 11:49
  • $\begingroup$ I am afraid there is no mistake in the question. Can you adapt your answer to the question? $\endgroup$
    – user505183
    Mar 6, 2019 at 11:54
  • $\begingroup$ In that case the square root will give us an imaginary number. Then, depending on the value of $a$ and $r$ you will also get an zeroth order contribution. One sec. $\endgroup$ Mar 6, 2019 at 12:22
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$\textbf{Full Edition.}$

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

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.

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  • $\begingroup$ Ready. Waiting for comments $\endgroup$ Mar 12, 2019 at 20:50
  • $\begingroup$ Take $r = 0$ and $a = 5$. Then equation 6, describing $t_0$ will give an imaginary result. Therefore the expression for $v(\epsilon)$ will give imaginary terms - this should not be the case. I am sorry but I am unable to (quickly) see where the error is - your method (and the notation) is impressive but hard to follow for me. $\endgroup$ Mar 12, 2019 at 22:15
  • $\begingroup$ @PiotrBenedysiuk $r=0$ eliminates $\varepsilon$ at all. $\endgroup$ Mar 12, 2019 at 22:45
  • $\begingroup$ It doesn't in the expression for your $v$. But the argument holds to if $r=1$ and $a=5$. $\endgroup$ Mar 12, 2019 at 23:05
  • $\begingroup$ @PiotrBenedysiuk Then should $6\ge|z|\ge4,,$ and I cannot find solution of $(1)$ $\endgroup$ Mar 12, 2019 at 23:40

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