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I have the following function that maps an ellipse in $z$-plane into a unit circle in $\zeta$-plane:

$$z = \sum_{n=0}^{N} \alpha_n\zeta^{n+1}$$

where $\alpha_n = a_n + ib_n$ is a constant, and $\zeta = e^{i\theta} = \cos\theta + i\sin\theta$ where [$\theta = 0...2\pi$].

I managed to obtain the constant $\alpha_n$ through iterative processes to transform an ellipse into a unit circle. The result is shown as the blue curve in the plot.enter image description here where the following constants $\alpha_n$ has been used to transform the ellipse into a unit circle (blue curve in the figure),

1.5000 + 0.0000i -0.0000 - 0.0000i -0.5000 + 0.0000i -0.0000 + 0.0000i 0.0000 + 0.0000i 0.0000 + 0.0000i 0.0000 - 0.0000i -0.0000 + 0.0000i -0.0000 + 0.0000i 0.0000 - 0.0000i -0.0000 - 0.0000i -0.0000 + 0.0000i 0.0000 + 0.0000i 0.0000 + 0.0000i 0.0000 - 0.0000i 0.0000 - 0.0000i -0.0000 - 0.0000i -0.0000 - 0.0000i -0.0000 + 0.0000i

Now, my question is, how to transform $z = \sum_{n=0}^{N} \alpha_n\zeta^{n+1}$ into another function as below,

$$z = \alpha_0\zeta + \sum_{n=1}^{N} \dfrac{\alpha_n}{\zeta^n}$$

Note that the series in the second function $n$ is started from $1$.

The goal is to obtain the constant $\alpha_0$ and $\alpha_n$ of the second function as such it will also produce a transformation from an ellipse into a unit circle.

Can anyone help me?

PS. I am thinking of using Fourier series/transform? but I don't really know how to implement them

BeeTiau
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  • It seems that $\alpha_n=0$ for $n>0$, so the series is already in the form you wanted, or am I missing something here? – caverac May 23 '18 at 10:28
  • Hi @caverac, the second function differs with the second one, i.e. the series of the second in the inverse of the first. – BeeTiau May 23 '18 at 10:31
  • Right, but if the coefficients are all zero, except for $n=0$ then then the two expressions are the same, right? – caverac May 23 '18 at 10:33
  • @caverac, the coefficient $\alpha_n$ cannot be zero since it is needed to conformally map the ellipse into a unit circle. Basically, if $\alpha_n=0$ the two are similar, as you said, and both will map a unit circle in $z$-plane into a unit circle in $\zeta$-plane. – BeeTiau May 23 '18 at 10:36
  • Then I’m confused by the numbers you put in your question. All of them seem to be zero, except for the first one – caverac May 23 '18 at 10:39
  • Ah, I see. Thanks @caverac. Those values are obtained from an iterative process and only the first and the third are non-zero, i.e. $\alpha_1 = \alpha_3 \neq 0$ – BeeTiau May 23 '18 at 10:41
  • Even If $\alpha_n \ne 0$ for $n > 0$, $\zeta^{-n}$ is just the complex conjugate of $\zeta^n$. So I'm guessing conjugation of $\alpha_n$ may be all you need. – Andy Walls May 23 '18 at 10:42
  • @AndyWalls, I didn't get what you mean. How do we get the constants $\alpha_n$ of the second function as such it will also map an ellipse into a unit circle? – BeeTiau May 23 '18 at 10:46
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    I'm speculating that the $\alpha_n$ of the second function are just the complex conjugates of the $\alpha_n$ of the first function. – Andy Walls May 23 '18 at 10:51
  • I haven't tried that, and will try that soon. But my guess it will not be the case, because the conjugate of $\alpha$ will give us the same value since, according to the data I gave in the post, the imaginary part is all zero. But I will still try this... – BeeTiau May 24 '18 at 17:10
  • does anyone else have any other idea how to solve this problem? – BeeTiau May 25 '18 at 17:44

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