Find the real zeros of a complex equation

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  • #1
joshmccraney
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Hi PF!

Here's my equation
Code:
0.5 Sech[0.997091 Sqrt[
   2.4674 + \[Xi]\[Xi]^2]] (-1. ((0. +
        3.7011 I) + (0. + 1.5 I) \[Xi]\[Xi]^2) Cosh[
     0.997091 Sqrt[2.4674 + \[Xi]\[Xi]^2]] -
   1. \[Sqrt](((0. + 3.7011 I) + (0. + 1.5 I) \[Xi]\[Xi]^2)^2 Cosh[
         0.997091 Sqrt[2.4674 + \[Xi]\[Xi]^2]]^2 +
       4. (2.4674 + \[Xi]\[Xi]^2)^(3/2)
         Cosh[0.997091 Sqrt[2.4674 + \[Xi]\[Xi]^2]] Sinh[
         0.997091 Sqrt[2.4674 + \[Xi]\[Xi]^2]]))
Making it real, it plots well, but I cannot identify the first zero of the real component of the equation. NSolve and Reduce fail, though it plots so clearly.

Thanks for your time!
 
Last edited:

Answers and Replies

  • #2
20,516
4,394
Hi PF!

Here's my equation
Code:
0.5 Sech[0.997091 Sqrt[
   2.4674 + \[Xi]\[Xi]^2]] (-1. ((0. +
        3.7011 I) + (0. + 1.5 I) \[Xi]\[Xi]^2) Cosh[
     0.997091 Sqrt[2.4674 + \[Xi]\[Xi]^2]] -
   1. \[Sqrt](((0. + 3.7011 I) + (0. + 1.5 I) \[Xi]\[Xi]^2)^2 Cosh[
         0.997091 Sqrt[2.4674 + \[Xi]\[Xi]^2]]^2 +
       4. (2.4674 + \[Xi]\[Xi]^2)^(3/2)
         Cosh[0.997091 Sqrt[2.4674 + \[Xi]\[Xi]^2]] Sinh[
         0.997091 Sqrt[2.4674 + \[Xi]\[Xi]^2]]))
Making it real, it plots well, but I cannot identify the first zero of the real component of the equation. NSolve and Reduce fail, though it plots so clearly.

Thanks for your time![/CODE]
Can you please write this out using LaTex, and also provide the plot you obtained?
 
  • #3
joshmccraney
Gold Member
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Can you please write this out using LaTex, and also provide the plot you obtained?
Yea, I left it like this because it should be easy to plot in Mathematica via copy-paste. However, here's a much simpler version:

Code:
a = -5;
c = 1;
sol[bb_] := x /. NSolve[a x^2 + b I x + c == 0, x]
Plot[{Re[sol[bb]], Im[sol[b]]}, {b, -10, 10}, PlotRange -> All]
The equation is ##ax^2+b i x + c##. The plot attached (yellow is imaginary and blue is real solution).

I don't know how to locate the zeros numerically in Mathematica. Any idea?
 

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  • #4
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This is probably a stupid question, but why can't you just use the quadratic formula?
 
  • #5
joshmccraney
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This is probably a stupid question, but why can't you just use the quadratic formula?
It's not stupid, but Mathematica does not give me an output when I ask for the real part of the solution. Example:

Code:
sol[b_] := (-b I \[PlusMinus] Sqrt[-b^2 - 4 a c])/(2 a)
Re[sol[1]]
it outputs ##-Re[-i \pm \sqrt{19}]/10## (using values listed above).
 
  • #6
DrClaude
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Calling f[x] the function you defined in the OP, try
Code:
FindRoot[Re[f[x]], {x, 0.4}]
 
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  • #7
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Solving it using the quadratic formula, I get $$x=\frac{bi\pm \sqrt{20-b^2}}{10}$$
What's wrong with that?
 
  • #8
joshmccraney
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The FindRoot did not work for me, but I appreciate your help Dr Claude. Chet, The issue is having Mathematica grab the real and imaginary component of the quadratic solution. What I ended up doing is taking the determinant and finding the zeros of this. Turns out this is equivalent to what I'm looking for.

Thanks for the help everyone!
 
  • #9
DrClaude
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The FindRoot did not work for me
I'm confused. I copied and pasted your equation in #1 and it worked fine.
 
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  • #10
joshmccraney
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I'm confused. I copied and pasted your equation in #1 and it worked fine.
Oops, I must have copied something wrong, because now it does in fact work. Thanks!
 
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