<s>[INST] Defining the equation of an ellipse in the complex plane 
Usually the equation for an ellipse in the complex plane is defined as $\lvert z-a\rvert + \lvert z-b\rvert = c$ where $c>\lvert a-b\rvert$. If we start with a real ellipse, can we define it in the manner below?


For $x,y,h,k,a,b\in\mathbb{R}$ such that $a,b\neq 0$, we define a real ellipse as
$$
  \frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1.
$$
Let $z = \frac{x}{a} + i\frac{y}{b}$ and $z_0 = \frac{h}{a} + i\frac{k}{b}$. If we expand the equation for an ellipse, we have
$$
  \frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{h^2}{a^2} + \frac{k^2}{b^2} -
  \frac{2xh}{a^2} - \frac{2yk}{b^2} = 1.
$$
Notice that $\lvert z\rvert^2 = \frac{x^2}{a^2} + \frac{y^2}{b^2}$ and $\lvert z_0\rvert^2 = \frac{h^2}{a^2} + \frac{k^2}{b^2}$. Now, let's write the ellipse as
$$
  \lvert z\rvert^2 + \lvert z_0\rvert^2 - \frac{2xh}{a^2} - \frac{2yk}{b^2} +
  \frac{yh}{ab}i - \frac{yh}{ab}i + \frac{xk}{ab}i - \frac{xk}{ab}i =
  \lvert z\rvert^2 + \lvert z_0\rvert^2 - \bar{z}z_0 - z\bar{z}_0 = 1.
$$
Thus, the equation of an ellipse in the complex plane is
$$
  (z - z_0)(\bar{z} - \bar{z}_0) = \lvert z - z_0\rvert^2 = 1\Rightarrow
  \lvert z - z_0\rvert = 1
$$
where $z$ and $z_0$ are defined above.
 [/INST] Your answer is correct, in the terms you are using.  But note that in the standard equation
$$|z-a|+|z-b|=c\ ,$$
it is implicitly assumed that $z=x+iy$.  Since your $z$ has a different form from this, you have obtained a different answer.
If you wish to plot your equation in the complex plane, you will have to rescale the axes and so the graph of $|z-z_0|=1$, which is "usually" a circle, will be "squeezed" so that it becomes an ellipse.
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