How to demonstrate that an elliptical circulation distribution can be induced by an elliptical chord distribution?
First, let's transform the coordinates to polar coordinates for simplicity by introducing a variable $\theta$ such that:
$$z=-s\cos{\theta}$$
In this coordinate, the chord distribution is then:
$$c(\theta)=c_C\sin{\theta}$$
Let's write the unknown lift distribution of the bound vortex as a Fourier sine series, again in the $\theta$ coordinate:
$$\Gamma(\theta) = \sum_{n=1}^{\infty}A_n\sin{n\theta}$$
Through Biot-Savart, the induced AOA can be computed from the trailing vortices, such that:
$$\alpha_i(z) = \frac{1}{4\pi V_\infty}\int_{-s}^{s}{\frac{(d\Gamma/d\zeta)}{z-\zeta}d\zeta} = \frac{1}{4\pi V_\infty}\int_{0}^{\pi}{\frac{\sum_{n=1}^{\infty}{nA_n\cos{n\theta_0}}}{\cos{\theta_0}-\cos{\theta}}d\theta_0} = \frac{1}{4V_\infty}\sum_{n=1}^{\infty}{nA_n\frac{\sin{n\theta}}{\sin{\theta}}}$$
Now collecting everything together and put into your original equation, we have:
$$\Gamma(\theta) = \frac{1}{2}a_{2D}V_{\infty}c_C\sin{\theta} \left[ \alpha - \beta - \frac{1}{4V_\infty}\sum_{n=1}^{\infty}{nA_n\frac{\sin{n\theta}}{\sin{\theta}}} \right] = \sum_{n=1}^{\infty}A_n\sin{n\theta}$$
The key, at this point, is to realize that each $\frac{\sin{n\theta}}{\sin{\theta}}$ is linearly independent to the others, so in order for the other terms to satisfy the above relation for all $\theta$, they must be simultaneously zero. That is,
$$A_n = \begin{cases}
      A_1 &amp; n=1\\
      0 &amp; n&gt;1\\
    \end{cases}$$
So now the circulation and chord distributions have the same form, which are elliptical.