# Find the real zeros of a complex equation

• Mathematica
Gold Member
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

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Chestermiller
Mentor
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?

Gold Member
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?

#### Attachments

• 14.9 KB Views: 32
Chestermiller
Mentor
This is probably a stupid question, but why can't you just use the quadratic formula?

Gold Member
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).

DrClaude
Mentor
Calling f[x] the function you defined in the OP, try
Code:
FindRoot[Re[f[x]], {x, 0.4}]

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

Gold Member
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
Mentor
The FindRoot did not work for me
I'm confused. I copied and pasted your equation in #1 and it worked fine.

joshmccraney
Gold Member
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!

DrClaude