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Does anyone know any Approximation methods to find complex zeros?

  1. Jul 24, 2006 #1
    Does anyone know any relatively simple approximation methods for approximating complex zeros of an analytic function, like say, cos(1/z), for example?


    Edwin G. Schasteen
  2. jcsd
  3. Jul 24, 2006 #2


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    For that particular example, all the zero of the function are real since for real x,y we have

    [tex]\cos (x+iy) = \cos (x) \mbox{cosh} (y) - i\sin (x) \mbox{sinh} (y)[/tex]

    so that upon setting [tex]\cos (x+iy) = 0[/tex] and equating the real and imaginary parts we require that the system of equations

    [tex]\cos (x) \mbox{cosh} (y)=0, \, \, \sin (x) \mbox{sinh} (y)=0[/tex]

    hold. The second of these equations is zero if and only if either y=0 or [tex]x=k\pi, k=0, \pm 1, \pm 2,\ldots ,[/tex] and the first equation cannot equal zero if [tex]x=k\pi, k=0, \pm 1, \pm 2,\ldots ,[/tex] so we must require that y=0 in order to satisfy the second equation and also require that [tex]\cos (x) =0[/tex] to make the first equation hold. Hence the only values of z such that [tex]\cos (z)=0[/tex] are [tex]z=\frac{\pi}{2}+k\pi, k=0, \pm 1, \pm 2,\ldots .[/tex]
  4. Jul 25, 2006 #3


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    Except that "that particular example" was [itex]cos(\frac{1}{z})= 0[/itex], not cos(z)= 0!

    Newton's method works nicely for complex values as well as real values: choose some starting [itex]z_0[/itex] and iterate
    [tex]z_{n+1}= z_n- \frac{f(z_n)}{f '(z_n)}[/tex]
  5. Jul 25, 2006 #4
    yes, but 1/z is real iff z (neq 0) is real
  6. Jul 25, 2006 #5
    Thanks! Newton's method would be perfect then!

    Benorin has a good point about Cos(1/z)=0 having no complex roots. That function was a poor example, a better choice would have been
    cos(1/z) - 2 = 0, which I believe, has all complex roots, except for the essential singularity at the origin. Is this correct? One more question....how come when you shift cos(1/z) vertically by some real valued amount k, cos(1/z)+k, the simple poles move off into the complex plane but the essential singularity at the origin never seems to move no matter how big k is? What, if anything, happens to the essential singularity as k approaches positive infinity?

    It seems that poles of finite order can be moved about in the complex plane by vertically shifting the function by adding or subtracting a real valued constant, and by shifting the function horizontally by adding or subtracting the independent variable z by some constant k,
    f(z-k). It seems that the only way to shift the essential singularity is by adding or subtracting the independent variable z by some constant k. Is this correct?


  7. Jul 25, 2006 #6
    yes this is correct (the all complex roots part - a singularity isn't a root!) because [itex]|\cos{z}| \leq 1[/itex] for all real z. The reason that you can't move the singularity by shifting vertically is that the singularity comes from the 1/z inside the cosine. So, for example, cos(1/(z-k)) will have a singularity at z=k (because 1/(z-k) has a singularity at z=k). Also, [itex]\cos(1/z)[/itex] is analytic for all [itex]z \neq 0[/itex] so I don't know what other poles you're referring to later in your post.

    Certainly if f(z) has a finite number of poles (of any order) then f(z-k) will have the same number, just shifted by k. And f(z) will never have the same roots as f(z)-k for [itex]k \neq 0[/itex] (to find the roots of f(z) you solve f(z)=0, and to find the roots of f(z)-k you solve f(z)=k). Note that a root is not the same as a "pole of finite order." Vertically shifting a function is never going to change its poles (in other words, for any k, f(z) is analytic iff f(z)+k is analytic).
    Last edited: Jul 25, 2006
  8. Jul 26, 2006 #7
    Code (Text):
    Also, is analytic for all so I don't know what other poles you're referring to later in your post.
    I apologize about that. I was thinking about zeros but my interest is in the poles of a function similar to 1/[cos(1/z)-k]. I meant to say zeros
    cos(1/z) - k, but had "poles" on the brain :smile: Also, I noticed that
    1/[f(z) - k] is not vertically shifted, g(z) = 1/f(z) - k would be vertically shifted, but, as you pointed out, that would not move the poles of g. Your posts answer my question; thanks!

    Another question I have is as follows:

    Suppose you have an analytic function f(z) -k, like cos(1/z) - k, that is analytic everywhere except the origin. If you were to vertically shift such a function continuously with time, must zeros of a function, that is analytic in some open disc, that move off into the complex plane move continuously along some continuous arc, or is it possible for the zeros of the "locally" analytic function to "jump about" erratically in the complex plane as k is varied continuously with time?


    Last edited: Jul 26, 2006
  9. Jul 29, 2006 #8
    -Hallsoftivy..how can Newton-method like for complex numbers when you can't "order" them?..for example for real numbers a,b,c,d,e,f,..you can put a>b>c>d>e>... but this doesn't work for complex numbers.

    -Perhaps the best method to find complex roots is to consider as pointed by others the function:

    [tex] f(x+iy)=0=U(x,y)+iW(x,y)=0 [/tex] and consider it as a system with 2 unknown values (x,y) to be determined... knowing that these values must satisfy [tex] U(x,y)=W(x,y)=0 [/tex]
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