Complex root of nonlinear complex equation containing besselh

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Liber-T
Liber-T 2011 年 6 月 7 日
I need to find the complex number which will make the real part of the function and the imaginary part, both equals 0.
I have tried fsolve, cxroot(which is a function from this site) and an algorithm of my own based on elemenantary numerical analysis, but none of them works.
Something more, my equations contains besselh(0 and 1,1,k*x) where k is a constant and x a variable. x must be numerical to use besselh on matlab.
I only need one root, and I approximatively know where it is. Is there a solution?
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Liber-T
Liber-T 2011 年 6 月 13 日
There seems to be a mistake in my formula then, but the formula comes from a determinant, and I've heard that I could directly solve the determinant by using det. But when I read in matlab's help, there is no clue on how to use det that way.
Walter Roberson
Walter Roberson 2011 年 6 月 13 日
I had Maple search for a complex zero in the indicated range. After a number of hours of calculation it indicated that there is no zero there.
Matt, the function does not oscillate wildly, but it has a very steep gradient -- much too steep for any potential zero to be detected by the default number of linspace() steps.
Liber-T: if the formula comes from a determinant, then that would seem to imply that you are looking for places where the original matrix is *not* invertible ? If so then possibly there are better ways to calculate that.

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