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I have following equation $\frac{x^2}{\text{R}}+\frac{y \log \left(\frac{i x-i y+1}{-i x-i y+1}\right)}{2 x \left(1+\frac{i \log \left(\frac{i x-i y+1}{-i x-i y+1}\right)}{2 x}\right)}+1 = 0$

1 + x^2/R + (y Log[(1 + I x - I y)/(1 - I x - I y)])/(
 2 x (1 + (I Log[(1 + I x - I y)/(1 - I x - I y)])/(2 x))) = 0

where $R$ is taken to be $0.5$ , $x>0, Re@y >0 , Im@y<0$.

I want to solve this equation with x as parameter and plot in complex plane of y curve. Analytic analysis gives that at $x \to 0$ $y-> - i R/3$ and when $x \to \infty$ $y \to \sqrt{R/3}$. However i have problems since FindRoot chooses incorrect solution.

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    $\begingroup$ Never use upper-case letters (e.g., R) for variables as they can conflict with the internal function and variable names in Mathematica. And why did you put in R as a parameter when you know its value is 0.5?? $\endgroup$ Commented Apr 26 at 0:57
  • $\begingroup$ I am planning later to slightly chane it, so i leave place to change it. $\endgroup$ Commented Apr 26 at 8:10
  • $\begingroup$ Your expression uses the principal-valued $\text{Log}$ function. Is it possible there are no solutions using this branch? Can you find any solution using FindRoot when back-substituted gives a residue close to the accuracy of the root say $10^{-10}$ or so if using default WorkingPrecision? $\endgroup$ Commented Apr 26 at 15:37
  • $\begingroup$ Xian-Zu: First: You simply don't understand. You should never use a term like "R" for a variable or constant because it can conflict with Mathematica's internal name space. Use "r" instead. What if you used "N" for a variable? That is the numerical value of an expression. What a mess! $\endgroup$ Commented Apr 26 at 16:45

1 Answer 1

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Multiply by denominators and define

f[x_, y_] := (1 + 2 x^2) (2 x +  I Log[(1 + I x - I y)/(1 - I x - I y)]) + 
              y Log[(1 + I x - I y)/(1 - I x - I y)]

Now you see the structure by

 ComplexPlot3D[f[1, y], {y, -40 - 12 I, 40 + 12 I}, 
                PlotRange -> {0, 2 \[Pi]}]

Complex Plot abs and arg by color

The continuous coloring outside the singular region indicates absence of branch cuts to infinity.

There is a branch cut connecting two singular points

 ComplexPlot3D[f[1, y], {y, -1.3 - 2.12 I, 1.3 + 0.5 I}, 
         PlotRange -> All, ImageSize -> Small, Boxed -> False]

Zoom into ComplexPlot3D

Somewhere near to the cut there is a zero, a cone of the abs with a full range of 2 $\pi$ of colours of arg around the zero point.

  ComplexPlot3D[f[1, y], {y, -0.2 -  I, 0.2 - 0.4 I}, PlotRange -> All, 
     ImageSize -> Small, Boxed -> False]

quadratic minimum

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  • $\begingroup$ The root appears to be pure imaginary: The code theY = y /. FindRoot[f[1, y] == 0, {y, Rationalize[-0.771627 I, 0]}, WorkingPrecision -> 100] can be used to compute it to thousands of digits of accuracy. Back-substitution leads to a residual comparable to the precision of the root. $\endgroup$ Commented Apr 26 at 23:21

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