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Asked by Ed on 10 Feb 2013

Hello,

I am trying to numerically solve a nonlinear complex equation and I would like to find all the complex roots. The equation is of the type:

cot(z)*z = 1-z^2*(1+i*z)

Does a specific function exist to find all the complex roots or do I need to separate z in the real and imaginary parts?

Thanks in advance for your help!

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Answer by Matt J on 10 Feb 2013

Edited by Matt J on 10 Feb 2013

Accepted answer

If you have the Symbolic Math Toolbox, I think SOLVE can be used to get complex-valued solutions. For the numerical solvers, I'm pretty sure you do have to reformulate the problem in terms of real and complex parts. Also, I've never heard of a numerical solver that will robustly find multiple roots for anything except polynomials.

Matt J on 10 Feb 2013

However, you don't have to explicitly/analytically rewrite your equation to pose it in terms of real and complex parts. Here's what you can do instead,

f=@(z)cot(z)*z -(1-z^2*(1+i*z)); c=@(x) complex(x(1),x(2)); g=@(x) abs(f(c(x)));

>> sol=c(fminsearch(g,[1;1])) %one root

sol =

-1.0900e-05 - 3.4270e-05i

Ed on 10 Feb 2013

Thanks for answering, I used your code and everything is fine, but I still have a question (sorry!):

the root that you find with the second method is just one. What should I do to find the others? Do I have to change the interval of the search? If so, this method should be much easier than to pose the equation in terms of real and complex parts!

Thanks again!

Matt J on 10 Feb 2013

FMINSEARCH doesn't give you control over the interval of the search. It just let's you choose a starting point hopefully close to the thing you're trying to find. If you know the approximate locations of the other roots, you could try initializing at them. Beyond that, see my earlier remark " **I've never heard of a numerical solver that will robustly find multiple roots for anything except polynomials.** "

Answer by Azzi Abdelmalek on 10 Feb 2013

Use fzero function

doc fzero

f=@(z)cot(z)*z -(1-z^2*(1+i*z)) z0=i; sol=fzero(f,z0); % the solution is near z0

Matt J on 10 Feb 2013

It's interesting that this worked for z0=i, but it appears to be just a fluke. FZERO can't really handle complex-valued functions. Note,

>> sol=fzero(f,1+i) Error using fzero (line 309) Function value at starting guess must be finite and real.

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