Find the real zeros of a complex equation

[member 428835]

Hi PF!

Here's my equation

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!

 

[Chestermiller]

Hi PF!

Here's my equation

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?

 

[member 428835]

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:

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?

 

[Chestermiller]

This is probably a stupid question, but why can't you just use the quadratic formula?

 

[member 428835]

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:

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).

 

[DrClaude]

Calling f[x] the function you defined in the OP, try

FindRoot[Re[f[x]], {x, 0.4}]

 

[Chestermiller]

Solving it using the quadratic formula, I get $$x=\frac{bi\pm \sqrt{20-b^2}}{10}$$
What's wrong with that?

 

[member 428835]

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!

 

[DrClaude]

The FindRoot did not work for me

I'm confused. I copied and pasted your equation in #1 and it worked fine.

 

[member 428835]

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!