The hyperbola
$$x^2 - y^2 = 1$$
has a simple expression in the complex plane as $\{z^2 + \bar{z}^2 = 2\}$.
Is there a similarly simple expression for a hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$? Or an ellipse?
I know we can express hyperbolae and ellipses as images of vertical and horizontal lines under the sine function.
Well, there's one way to find out. Solving the system $$z = x + iy$$ $$\overline{z} = x - i y$$ for $x$ and $y$, we get $$x = \frac{z+\overline{z}}{2}$$ and $$y = \frac{z-\overline{z}}{2i}.$$ Plugging these into the equation $$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1,$$ we get $$\frac{(z+\overline{z})^2}{4a^2} + \frac{(z-\overline{z})^2}{4b^2}=1.$$ If there's no relation between $a$ and $b$ then the $z \overline{z}$ term won't cancel like it did in the $a = b$ case you gave above.
However, if instead of starting with an equation like $x^2/a^2 - y^2/b^2 = 1$ you think more geometrically there's a better way of doing things. If we let $p$ and $q$ be the foci of a hyperbola, we can write an equation like $$||z - p| - |z - q|| = 2a$$ which simply encodes the geometric definition of the hyperbola.