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Homework Help: How do you always put a complex function into polar form?

  1. Dec 20, 2017 #1
    1. The problem statement, all variables and given/known data
    It's not a homework problem itself, but rather a general method that I imagine is similar to homework. For a given elementary complex function in the form of the product, sum or quotient of polynomials, there are conventional methods for converting them to polar form. The problem however is that only very few people across varying sources understand the subject well enough to explain the concise procedure of converting these functions to polar form, even for specific cases, and thus the commonality between the solutions of different problems is convoluted.

    2. Relevant equations
    What is the procedure for representing elementary functions of a complex variable in polar form by finding their absolute value and argument?

    3. The attempt at a solution
    From what I have seen, I can guess that the first step is likely to arrange the function into the form of f(z) = u +iv which can help with both finding the absolute value and the argument, though I won't say I know that for certain. Beyond that, I am not sure of a definite pattern, although the atan2 operator comes up often which makes sense as the angle of a complex number could be found with an arctangent.
     
    Last edited: Dec 20, 2017
  2. jcsd
  3. Dec 20, 2017 #2

    Orodruin

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    Without getting more specific, ##|f(z)|e^{i\arg(f(z))}## is about as general you can get (don't forget the ##i##!), it is what it is.

    ##\LaTeX## seems to be working fine. Note that MathJax requires different delimiters than regular ##\LaTeX##, see https://www.physicsforums.com/help/latexhelp/.
     
  4. Dec 20, 2017 #3
    Okay but I'm not asking if something is as general as it can get, I'm asking for the specific procedure for writing a complex function in polar form, most often for algebraic and transcendental functions.
     
  5. Dec 20, 2017 #4

    Orodruin

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    You take the absolute value and the argument and put it on the given form. Your question in the end boils down to "how do I take the absolute value and argument" to which the answer would generally be "you compute the corresponding functions".
     
  6. Dec 20, 2017 #5
    You keep making claims but you don't seem to actually have evidence to support them. Given the lack of procedure after I asked for it twice, it seems you do not have enough experience in this subject to address the problem. Please refrain from interfering with this thread further as your posts are not beneficial. There are very clear and concise produces for this type of problem just as there are very clear and precise produces for other general problems like finding the 0s of elementary and algebraic functions.

    In case you have not seen, I opened another thread which asks a more specific question focused only on polynomials. I would prefer this thread to be deleted so as to not detract focus from the other.
     
    Last edited: Dec 20, 2017
  7. Dec 20, 2017 #6

    Orodruin

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    Really? I would say you are the one being vague in your question. Putting any complex number on polar form is equivalent to computing its magnitude and argument - there is nothing more to it, it is a well defined procedure. That you fail to specify a reasonable question based on this while thinking that you have is not my problem.

    Based on your very vague questioning and clear lack of appreciation of the fact that the problem lies with your question, I suggest that you read up on basic complex analysis.
     
  8. Dec 20, 2017 #7
    If it's such a "well defined" procedure as you insist, then why do you still avoid writing it out and instead derail the thread with your arbitrary opinions? If the question isn't reasonable, how do you understand it well enough to know it is a well defined procedure? This is math, not a political poll, you're clearly not contributing to the topic that would benefit people to understand as proven by the fact that the subject is taught in universities worldwide for practical applications.
    If you find the generality unclear or difficult to work with, here is a common method of constructively asking for clarity instead of rudely imposing arbitrary opinions to derail a topic:
    "The procedure varies widely with different groups of functions. Can you please specify a single group of functions to work with: e.g. polynomials?"
     
    Last edited: Dec 20, 2017
  9. Dec 20, 2017 #8

    fresh_42

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    Maybe you were looking for
    ##|f(|z|e^{iarg(z)})|e^{iarg(f(|z|e^{iarg(z}))}##
     
  10. Dec 20, 2017 #9
  11. Dec 20, 2017 #10

    Mark44

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    ##re^{i\theta}## is the polar form of z = a + bi. In the polar form, ##r = \sqrt{a^2 + b^2}## and ##\theta = \arctan(\frac b a)##. There's not a lot more to say.

    In your first post, you were asking about complex functions, not complex numbers, so by not posting a clearer question, you didn't get a clear answer.
     
  12. Dec 20, 2017 #11
    Well I don't have the means of ruling out that the principals of representing complex numbers in polar form don't also apply to complex functions if that is in fact the case. For instance, extrapolating from what you presented, if you have a complex function ##w = u(x,y) + iv(x,y)##, is it true that $$r = \sqrt{ww*}?$$
     
  13. Dec 20, 2017 #12

    Mark44

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    Yes
     
  14. Dec 20, 2017 #13
    That is helpful to know, thank you for confirming that. For the case of ## \theta = arctan( v / u)##, it doesn't seem like that generally simplifies to anything nice in order to tell you what theta is conventionally, like for instance if you tried to plug that in for w = z^2 where u+iv = x^2-y^2+2ixy. So does that mean the more conventional route is to simply substitute ##z## for ##re^{i \theta}## and, I don't know, hope it simplifies to an easy exponential form?
     
    Last edited: Dec 20, 2017
  15. Dec 20, 2017 #14

    Mark44

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    Conversion to polar form isn't a magic bullet. What you wrote above isn't crystal clear. If ##z = u + iv = x^2 - y^2 + 2ixy##, then ##z^2 = (x^2 - y^2 + 2ixy)^2##. This could be expanded without too much effort, but it's debatable whether putting it into polar form would be advantageous.
     
  16. Dec 21, 2017 #15

    kuruman

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    This particular function that you chose is simple and pleasing in polar form. If you define ##x=r \cos \theta## and ##y=r \sin \theta##, you get ##z=r^2 \cos(2\theta) +i \sin(2\theta)=(re^{i\theta})^2##. Then finding ##z^2## needs even less effort than expanding the polynomial.
     
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