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Z = u^v

  1. Aug 22, 2015 #1
    If I define the complex number z = r exp(i θ) how z = uv, so, how to express u and v in terms of r and θ?

    u(r, θ) = ?
    v(r, θ) = ?

    And the inverse too:

    r(u, v) = ?
    θ(u, v) = ?
  2. jcsd
  3. Aug 22, 2015 #2


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    Have you looked at this article?

  4. Aug 22, 2015 #3


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    I don't understand your question. You ask, apparently, for z in terms of "u" and "v" but have not said what "u" and "v" are! Are you referring to the representation of a complex function as "z(x+ iy)= u(x,y)+ iv(x,y)" where u and v are real valued function of the real variables x and y? If so then [itex]z= re^{i\theta}[/itex] is NOT "[itex]u^v[/itex]". [itex]z= re^{i\theta}= r(cos(\theta)+ isin(\theta)[/itex] so that [itex]u= r cos(\theta)[/itex] and [itex]v= r sin(\theta)[/itex].
  5. Aug 22, 2015 #4


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    I'm not sure what you are looking for, but [itex]u=re^{i\theta},\ v=1[/itex] works.
    However there are an infinite number of possibilities, by using [itex]u=r^ne^{ni\theta}\ and\ v=\frac{1}{n}[/itex].
  6. Aug 24, 2015 #5
    Z is a complex number, u is a real number and v is a real number too. Is just another way of express the complex numbers...

    So, is possible convert the expression z = x + i y in z = uv ? Is possible express u and v in terms of x and y?
  7. Aug 24, 2015 #6


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    Well, let's try. You know that [itex]z=x+iy [/itex] can also be expressed as [itex] z=re^{i\phi}[/itex], where [itex] r=\sqrt{x^{2}+y^{2}}[/itex] and [itex]\phi = \arcsin(\frac{y}{r}) [/itex]. Therefore, we obviously have [itex] z=e^{\ln r}\cdot e^{i\phi}= e^{\ln r + i\phi}[/itex]. ...
  8. Aug 24, 2015 #7
    Yeah! I thought this... but, I was unsatisfied com this 'conversion' and so I posted my doubt here because the most experiente could see something better...

    Anyway! Thank you!
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