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Mapping of multivalued complex function.

  1. Jan 20, 2010 #1
    A complex function [tex]f\left(z\right)=\sqrt{z}[/tex] can be splitted into two branches:
    1. Principal branch: [tex]f_{1}\left(z\right)=\sqrt{r} e^{i \left(\theta/2\right)}[/tex]
    2. Second branch: [tex]f_{2}\left(z\right)=\sqrt{r} e^{i \left[\left(\theta+2\pi\right) /2\right]}[/tex]

    My question is, is there a way to show/proof that the principal branch only map the z-plane to the right half plane of w-plane (Re w > 0) to which positive imaginary semiaxis is added, and that the second branch only map the z-plane to the left half plane of w-plane (Re w < 0) to which negative imaginary semiaxis is added??

    Then, how exactly is to plot these branches in z-plane and w-plane? I am confused, because r and theta could be anywhere in the z-plane, and hence it could be mapped to any point in the w-plane. Please tell me if what I said is wrong.

    Thanks in advance.
  2. jcsd
  3. Jan 20, 2010 #2
    if you parametrize the entire complex plane by [tex]( z=re^{i\theta} : r\in R, r\geq0, -\pi\leq\theta\leq\pi)[/tex]

    Then the range of [tex]\sqrt{r}[/tex] from r>0 remains [tex]r\in R, r\geq0 [/tex]
    But the range of [tex]\theta /2[/tex] from [tex]-\pi\leq\theta\leq\pi[/tex] is
    [tex]-\frac{\pi}{2}\leq \theta /2 \leq \frac{\pi}{2}[/tex]

    So the range of the first branch has a parametrization:
    [tex]( w=re^{i\theta}: r\in R, r\geq0, -\frac{\pi}{2}\leq \theta \leq \frac{\pi}{2} )[/tex]

    Which is off course a parametrization of the right half plane Re(w)>0
  4. Jan 26, 2010 #3
    I see it now. Thanks a lot.
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